10.4: The Case of H₂⁺ (2023)

One can develop an intuitive sense of molecular orbitals and what a chemical bond is by considering the simplest molecule, \(\ce{H_2^{+}}\). This ion consists of two protons held together by the electrostatic force of a single electron. Clearly the two protons, two positive charges, repeal each other. The protons must be held together by an attractive Coulomb force that opposes the repulsive Coulomb force. A negative charge density between the two protons would produce the required counter-acting Coulomb force needed to pull the protons together. So intuitively, to create a chemical bond between two protons or two positively charged nuclei, a high density of negative charge between them is needed. We expect the molecular orbitals that we find to reflect this intuitive notion.

The electronic Hamiltonian for \(\ce{H_2^{+}}\) is

\[\hat {H}_{elec} (r, R) = -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {e^2}{4 \pi \epsilon _0 r_A} - \dfrac {e^2}{4 \pi \epsilon _0 r_B} + \dfrac {e^2}{4 \pi \epsilon _0 R} \label {10.13}\]

where \(r\) gives the coordinates of the electron, and \(R\) is the distance between the two protons. Although the Schrödinger equation for \(\ce{H_2^{+}}\) can be solved exactly because there is only one electron, we will develop approximate solutions in a manner applicable to other diatomic molecules that have more than one electron.

For the case where the protons in \(\ce{H_2^{+}}\) are infinitely far apart, we have a hydrogen atom and an isolated proton when the electron is near one proton or the other. The electronic wavefunction would just be \(1s_A(r)\) or \(1s_B(r)\) depending upon which proton, labeled A or B, the electron is near. Here 1sA denotes a 1s hydrogen atomic orbital with proton A serving as the origin of the spherical polar coordinate system in which the position \(r\) of the electron is specified. Similarly \(1s_B(r)\) has proton B as the origin. A useful approximation for the molecular orbital when the protons are close together therefore is a linear combination of the two atomic orbitals. The general method of using

\[\psi (r) = C_A 1s_A (r) + C_B1s_B (r) \label {10.14}\]

i.e. of finding molecular orbitals as linear combinations of atomic orbitals is called the Linear Combination of Atomic Orbitals - Molecular Orbital (LCAO-MO) Method. In this case we have two basis functions in our basis set, the hydrogenic atomic orbitals 1sA and lsB.

For \(\ce{H_2^{+}}\), the simplest molecule, the starting function is given by Equation \(\ref{10.14}\). We must determine values for the coefficients, \(C_A\) and \(C_B\). We could use the variational method to find a value for these coefficients, but for the case of \(\ce{H_2^{+}}\) evaluating these coefficients is easy. Since the two protons are identical, the probability that the electron is near A must equal the probability that the electron is near B. These probabilities are given by \(|C_A|^2\) and \(|C_B|^2\), respectively. Consider two possibilities that satisfy the condition \(|C_A|^2 = |C_B|^2\); namely, \(C_A = C_B = C_{+} \text {and} C_A = -C_B = C_{-}\). These two cases produce two molecular orbitals:

\[\psi _+ = C_+(1s_A + 1s_B)\]

\[\psi _{-} = C_{-}(1s_A - 1s_B) \label {10.15}\]

The probability density for finding the electron at any point in space is given by \(|{\psi}^2|\) and the electronic charge density is just \(|e{\psi}^2|\). The important difference between \(\psi _+\) and \(\psi _{-}\) is that the charge density for \(\psi _+\) is enhanced between the two protons, whereas it is diminished for \(\psi _{-}\) as shown in Figures \(\PageIndex{1}\). \(\psi _{-}\) has a node in the middle while \(\psi _+\) corresponds to our intuitive sense of what a chemical bond must be like. The electronic charge density is enhanced in the region between the two protons. So \(\psi _+\) is called a bonding molecular orbital. If the electron were described by \(\psi _{-}\), the low charge density between the two protons would not balance the Coulomb repulsion of the protons, so \(\psi _{-}\) is called an antibonding molecular orbital.

Now we want to evaluate \(C_+\) and \(C_-\) and then calculate the energy. The bonding and antibonding character of \(\psi _+\) and \(\psi _{-}\) also should be reflected in the energy. If \(\psi _+\) indeed describes a bonding orbital, then the energy of this state should be less than that of a proton and hydrogen atom that are separated. The calculation of the energy will tell us whether this simple theory predicts \(\ce{H_2^{+}}\) to be stable or not and also how much energy is required to dissociate this molecule.

10.4: The Case of H₂⁺ (1)
10.4: The Case of H₂⁺ (2)

Exercise \(\PageIndex{1}\)

From the information in Figure \(\PageIndex{1}\) for \(\ce{H_2^{+}}\), calculate the difference in the electronic charge density (C/pm3) at a point halfway between the two nuclei for an electron in the bonding molecular orbital compared to one in the antibonding molecular orbital.

The constants \(C_+\) and \(C_-\) are evaluated from the normalization condition. Bracket notation, \(<|>\), is used in Equation \(\ref{10.16}\) to represent integration over all the coordinates of the electron for both functions \(\psi _+\) and \(\psi _-\). The right bracket represents a function, the left bracket represents the complex conjugate of the function, and the two together mean integrate over all the coordinates.

\[\int \psi ^*_{\pm} \psi _{\pm} d\tau = \left \langle \psi _{\pm} | \psi _{\pm} \right \rangle = 1 \label {10.16}\]

(Video) Week 11-Lecture 61 : Molecular Orbital Theory for H2+

\[\left \langle C_{\pm} [ 1s_A \pm 1s_B ] | C_{\pm} [ 1s_A \pm 1s_B ]\right \rangle = 1 \label {10.17}\]

\[|C_\pm|^2 [ (1s_A | 1s_A) + (1s_B | 1s_B) \pm (1s_B | 1s_A) \pm (1s_A | 1s_B)] = 1 \label {10.18}\]

Since the atomic orbitals are normalized, the first two integrals are just 1. The last two integrals are called overlap integrals and are symbolized by S and S*, respectively, since one is the complex conjugate of the other.

Exercise \(\PageIndex{2}\)

Show that for two arbitrary functions \(\left \langle \varphi _B | \varphi _A \right \rangle \) is the complex conjugate of \(\left \langle \varphi _A | \varphi _B \right \rangle \) and that these two integrals are equal if the functions are real.

The overlap integrals are telling us to take the value of lsB at a point multiply by the value of lsA at that point and sum (integrate) such a product over all of space. If the functions don’t overlap, i.e. if one is zero when the other one isn’t and vice versa, these integrals then will be zero. It also is possible in general for such integrals to be zero even if the functions overlap because of the cancellation of positive and negative contributions, as was discussed in Section 4.4.

If the overlap integral is zero, for whatever reason, the functions are said to be orthogonal. Notice that the overlap integral ranges from 0 to 1 as the separation between the protons varies from \(R = ∞\) to \(R = 0\). Clearly when the protons are infinite distance apart, there is no overlap, and when \(R = 0\) both functions are centered on one nucleus and \(\left \langle 1s_A | 1s_B \right \rangle\) becomes identical to \(\left \langle 1s_A | 1s_B \right \rangle\), which is normalized to 1, because then \(1s_A = 1s_B\).

With these considerations and using the fact that \(1s\) wavefunctions are real so

\[ \left \langle 1s_A | 1s_B \right \rangle = \left \langle 1s_B | 1s_A \right \rangle = S \label {10.19}\]

Equation \(\ref{10.18}\) becomes

\[|C_{\pm}|^2 (2 \pm 2S ) = 1 \label {10.20}\]

The solution to Equation \(\ref{10.20}\) is given by

\[C_{\pm} = [2(1 \pm S )]^{-1/2} \label {10.21}\]

The energy is calculated from the expectation value integral,

\[E_{\pm} = \left \langle \psi _{\pm} | \hat {H} _{elec} | \psi _{\pm} \right \rangle \label {10.22}\]

which expands to give

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\[E_{\pm} = \dfrac {1}{2(1 \pm s)} [ \left \langle 1s_A |\hat {H} _{elec} | 1s_A \right \rangle + \left \langle 1s_B |\hat {H} _{elec} | 1s_B \right \rangle \pm \left \langle 1s_A |\hat {H} _{elec} | 1s_B \right \rangle \pm \left \langle 1s_B |\hat {H} _{elec} | 1s_A \right \rangle ] \label {10.23} \]

Exercise \(\PageIndex{3}\)

Show that Equation \(\ref{10.22}\) expands to give Equation \(\ref{10.23}\).

The four integrals in Equation \(\ref{10.23}\) can be represented by \(H_{AA}\), \(H_{BB}\), \(H_{AB}\), and \(H_{BA}\), respectively. Notice that A and B appear equivalently in the Hamiltonian operator, Equation \(\ref{10.13}\). This equivalence means that integrals involving \(1s_A\) must be the same as corresponding integrals involving \(ls_B\), i.e.

\[ H_{AA} = H_{BB} \label {10.24}\]

and since the wavefunctions are real,

\[H_{AB} = H_{BA} \label {10.25}\]


\[ E_{\pm} = \dfrac {1}{1 \pm S} (H_{AA} \pm H_{AB}) \label {10.26}\]

Now examine the details of HAA after inserting Equation \(\ref{10.13}\) for the Hamiltonian operator.

\[H_{AA} = \left \langle 1s_A | - \dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {e^2}{4\pi \epsilon _0 r_A}| 1s_A \right \rangle + \dfrac {e^2}{4\pi \epsilon _0 R} \left \langle 1s_A | 1s_A \right \rangle - \left \langle 1s_A | \dfrac {e^2}{4 \pi \epsilon _0 r_B } | 1s_A \right \rangle \label {10.27}\]

The first term is just the integral for the energy of the hydrogen atom, \(E_H\). The second integral is equal to 1 by normalization; the prefactor is just the Coulomb repulsion of the two protons. The last integral, including the minus sign, is represented by \(J\) and is called the Coulomb integral. Physically \(J\) is the potential energy of interaction of the electron located around proton A with proton B. It is negative because it is an attractive interaction. It is the average interaction energy of an electron described by the 1sA function with proton B.

Now consider \(H_{AB}\).

\[H_{AB} = \left \langle 1s_A | - \dfrac {\hbar ^2}{1m} \nabla ^2 - \dfrac {e^2}{4\pi \epsilon _0 r_B}| 1s_B \right \rangle + \dfrac {e^2}{4\pi \epsilon _0 R} \left \langle 1s_A | 1s_B \right \rangle - \left \langle 1s_A | \dfrac {e^2}{4 \pi \epsilon _0 r_A } | 1s_B \right \rangle \label {10.28}\]

In the first integral we have the hydrogen atom Hamiltonian and the H atom function 1sB. The function lsB is an eigenfunction of the operator with eigenvalue EH. Since EH is a constant it factors out of the integral, which then becomes the overlap integral, S. The first integral therefore reduces to EHS. The second term is just the Coulomb energy of the two protons times the overlap integral. The third term, including the minus sign, is given the symbol \(K\) and is called the exchange integral. It is called an exchange integral because the electron is described by the 1sA orbital on one side and by the lsB orbital on the other side of the operator. The electron changes or exchanges position in the molecule. In a Coulomb integral the electron always is in the same orbital; whereas, in an exchange integral, the electron is in one orbital on one side of the operator and in a different orbital on the other side.

Using the expressions for \(H_{AA}\) and \(H{AB}\) and substituting into Equation \(\ref{10.26}\) produces:

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\[\begin{align} E_{\pm} &= \dfrac {1}{1 \pm S} \left[ (E_H + \dfrac {e^2}{4\pi \epsilon_0 R}) (1 \pm S ) + J \pm K \right] \label {10.29} \\[4pt] &= \underbrace{E_H}_{\text{H Atom Energy}} + \underbrace{\dfrac {e^2}{4\pi \epsilon _0 R}}_{\text{Proton-Proton repulsion}} + \underbrace{\dfrac {J \pm K}{1 \pm S}}_{\text{Bonding Energy}} \label {10.30} \end{align} \]

The difference in energies of the two states \(\Delta E_{\pm}\) is then:

\[\begin{align} \Delta E_{\pm} &= E_{\pm} - E_H \label {10.30B} \\[4pt] &= \dfrac {e^2}{4\pi \epsilon _0 R} + \dfrac {J \pm K}{1 \pm S} \label {10.31}\end{align} \]

Equation \(\ref{10.30}\) tells us that the energy of the \(\ce{H_2^{+}}\) molecule is the energy of a hydrogen atom plus the repulsive energy of two protons plus some additional electrostatic interactions of the electron with the protons. These additional interactions are given by

\[ \dfrac {J \pm K}{1 \pm S}\]

If the protons are infinitely far apart then only \(E_H\) is nonzero. To get a chemical bond and a stable \(\ce{H_2^{+}}\) molecule, \(\Delta E_{\pm}\) (Equation \ref{10.30B}) must be less than zero and have a minimum, i.e.

\[ \dfrac {J \pm K}{1 \pm S}\]

must be sufficiently negative to overcome the positive repulsive energy of the two protons

\[\dfrac {e^2}{4 \pi \epsilon _0R }\]

for some value of \(R\). For large \(R\) these terms are zero, and for small \(R\), the Coulomb repulsion of the protons rises to infinity.

Exercise \(\PageIndex{4}\)

Show that Equation \(\ref{10.13}\) follows from Equation \(\ref{10.26}\).

We will examine more closely how the Coulomb repulsion term and the integrals \(J\), \(K\), and \(S\) depend on the separation of the protons, but first we want to discuss the physical significance of \(J\), the Coulomb integral, and \(K\), the exchange integral.

Both \(J\) and \(K\) have been defined as

\[ J = \left \langle 1s_A | \dfrac {-e^2}{4 \pi \epsilon _0 r_B } |1s_A \right \rangle = - \int \varphi ^*_{1s_A} (r) \varphi _{1s_A} (r) \dfrac {e^2}{4 \pi \epsilon _0 r_B } d\tau \label {10.32}\]

\[ K = \left \langle 1s_A | \dfrac {-e^2}{4 \pi \epsilon _0 r_A } |1s_B \right \rangle = - \int \varphi ^*_{1s_A} (r) \varphi _{1s_B} (r) \dfrac {e^2}{4 \pi \epsilon _0 r_A } d\tau \label {10.33}\]

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Note that both integrals are negative since all quantities in the integrand are positive. In the Coulomb integral, \(e \varphi ^*_{1s_A} (r) \varphi _{1a_A} (r)\) is the charge density of the electron around proton A, since r represents the coordinates of the electron relative to proton A. Since rB is the distance of this electron to proton B, the Coulomb integral gives the potential energy of the charge density around proton A interacting with proton B. J can be interpreted as an average potential energy of this interaction because \(e \varphi ^*_{1s_A} (r) \varphi _{1a_A} (r)\) is the probability density for the electron at point r, and \(\dfrac {e^2}{4 \pi \epsilon _0 r_B }\) is the potential energy of the electron at that point due to the interaction with proton B. Essentially, \(J\) accounts for the attraction of proton B to the electron density of hydrogen atom A. As the two protons get further apart, this integral goes to zero because all values for rB become very large and all values for \(1/r_B\) become very small.

In the exchange integral, K, the product of the two functions is nonzero only in the regions of space where the two functions overlap. If one function is zero or very small at some point then the product will be zero or small. The exchange integral also approaches zero as internuclear distances increase because the both the overlap and the 1/r values become zero. The product \(e \varphi ^*_{1s_A} (r) \varphi _{1a_B} (r)\) is called the overlap charge density. Since the overlap charge density is significant in the region of space between the two nuclei, it makes an important contribution to the chemical bond. The exchange integral, \(K\), is the potential energy due to the interaction of the overlap charge density with one of the protons. While J accounts for the attraction of proton B to the electron density of hydrogen atom A, \(K\) accounts for the added attraction of the proton due the build-up of electron charge density between the two protons.

Exercise \(\PageIndex{5}\)

Write a paragraph describing in your own words the physical significance of the Coulomb and exchange integrals for \(\ce{H2^{+}}\).

Figure \(\PageIndex{2}\) shows graphs of the terms contributing to the energy of \(\ce{H_2^{+}}\). In this figure you can see that as the internuclear distance R approaches zero, the Coulomb repulsion of the two protons goes from near zero to a large positive number, the overlap integral goes for zero to one, and J and K become increasingly negative.

10.4: The Case of H₂⁺ (3)

Figure \(\PageIndex{3}\) shows the energy of \(\ce{H_2^{+}}\) relative to the energy of a separated hydrogen atom and a proton as given by Equation \(\ref{10.30}\). For the electron in the antibonding orbital, the energy of the molecule, \(E_H(R)\), always is greater than the energy of the separated atom and proton.

10.4: The Case of H₂⁺ (4)

For the electron in the bonding orbital, you can see that the big effect for the energy of the bonding orbital, E+(R), is the balance between the repulsion of the two protons \(\dfrac {e^2}{4 \pi \epsilon _0R }\) and \(J\) and \(K\), which are both negative. \(J\) and \(K\) manage to compensate for the repulsion of the two protons until their separation is less than 100 pm (i.e the energy is negative up until this point), and a minimum in the energy is produced at 134 pm. This minimum represents the formation of a chemical bond. The effect of S is small. It only causes the denominator in Equation \(\ref{10.30}\) to increase from 1 to 2 as \(R\) approaches 0.

For the antibonding orbital, \(-K\) is a positive quantity and essentially cancels \(J\) so there is not sufficient compensation for the Coulomb repulsion of the protons. The effect of the \(-K\) in the expression, Equation \(\ref{10.30}\), for \(E_-\) is to account for the absence of overlap charge density and the enhanced repulsion because the charge density between the protons for \(\psi _-\) is even lower than that given by the atomic orbitals.

This picture of bonding in \(\ce{H_2^{+}}\) is very simple but gives reasonable results when compared to an exact calculation. The equilibrium bond distance is 134 pm compared to 106 pm (exact), and a dissociation energy is 1.8 eV compared to 2.8 eV (exact).

Exercise \(\PageIndex{6}\)

Write the final expressions for the energy of \(\psi _-\) and \(\psi _-\), explain what these expressions mean, and explain why one describes the chemical bond in H2+and the other does not.

Exercise \(\PageIndex{7}\)

Figure \(\PageIndex{2}\) shows that \(S = 1\) and \(J = K =1\) hartree when \(R = 0\). Explain why \(S\) equals 1 and \(J\) and \(K\) equal -1 hartree when \(R = 0\).


What is the bond length of H2? ›

Bond Length. In Section 4.1, we stated that the covalent bond in the hydrogen molecule (H 2) has a certain length, about 7.4 × 10 11 meters (m) or 74 picometers (pm).

Is He2+ stable or unstable? ›

He2^+2 ion is more stable than He2 molecule.

What is the electron configuration of the H2+ ion? ›

In this case, there are six bonds between the protons in the nucleus of the H2+ ion. Therefore, the electron configuration of the H2+ ion is 1s22p6 6d5.

What is the bond order of H2+? ›

Assertion : Bond order of H2+ is 0.5. Reason : Electrons are removed from the antibonding molecular orbital from H2.

How many bonds can H2 form? ›

Each water molecule can form two hydrogen bonds involving their hydrogen atoms plus two further hydrogen bonds utilizing the hydrogen atoms attached to neighboring water molecules.

Which is more stable H2 or H2+? ›

Bond order for both H2+ and H2− is same but H2+ is more stable than H2− due to absence of electrons in anti-bonding orbital.

Why is H2 more stable than H2+? ›

Because H2- contains one electron in the antibonding orbital which results in repulsion and decrease the stability . on the other hand H2+ does not contain any electron in the antibonding molecular orbital.

Is H2 or He2+ more stable? ›

He2+2 ion is more stable than He2 molecule.

How many electrons are in H2+? ›

The number of electrons in the species H2+, H2 and O2+ are 1, 2 and 15 respectively.

Why is hydrogen H+? ›

A hydrogen atom is made up of a nucleus with charge +1, and a single electron. Therefore, the only positively charged ion possible has charge +1. It is noted H+.

Why does hydrogen form H+? ›

When the Hydrogen atom loses an electron all that is left is a proton. It becomes the positively charged hydrogen ion known as H+. This is the form of Hydrogen that produces the ATP enzyme that powers our cells and mitochondria. The H+ hydrogen ion is the basis of the pH scale.

How is H2+ formed? ›

Abstract. The molecular ion of dihydrogen (H2+) is produced by 1 eV collisions of protons (H+) on amorphous water ice surfaces.

What is the bond energy of H2 H2 H2? ›

The bond energy of H2 is 104.3 kcal mol^-1 .

Is H2+ and H2 paramagnetic? ›

For diamagnetic character, there should not be any unpaired electron in the molecules formation. Here, for H2 the number of unpaired electron is 0. But in all other case there's an unpaired electron. Let it be H2- ,H2+ or He2+ ; there's an unpaired electron i.e, they are paramagnetic.

Why type of bond is H2? ›

Covalent molecules made of only one type of atom, like hydrogen gas (H2), are nonpolar because the hydrogen atoms share their electrons equally.

Does H2 have a bond order of 2? ›

For H2, bond order = 1/2 (2-0) = 1, which means H2has only one bond. The antibonding orbital is empty. Thus, H2 is a stable molecule.

Can H2 form double bond? ›

It can form only a single ionic bond. It can form a single covalent bond.

Is h2 a single bond? ›

Dihydrogen is an elemental molecule consisting of two hydrogens joined by a single bond.

What kind of bond is h2 an example of? ›

Here is a list of molecules that exhibit hydrogen bonding: water (H2O): Water is an excellent example of hydrogen bonding.

Is h2 a strong bond? ›

hydrogen bonding, interaction involving a hydrogen atom located between a pair of other atoms having a high affinity for electrons; such a bond is weaker than an ionic bond or covalent bond but stronger than van der Waals forces.

Does H2 exist? ›

Hydrogen in molecular form exists as (H2) which means two atoms of hydrogen combine to form (H2) and hence it is a diatomic molecule.

Which is stable H or H2? ›

H2 molecule is more stable than He - H molecule.

What is the formula for bond order? ›

Bond order (B.O.) is defined as one-half the difference between the number of electrons present in the bonding and the anti-bonding orbitals. Bond order (B.O) = 1 2 ( Number of bonding electrons - Number of the anti-bonding electron) Bond order = 1 2 (Nb- Na)

In which of the following H2 bond is strongest? ›

The strongest bond is F-H-F. The interaction between the electronegativity of the bound atom and hydrogen determines the strength of the hydrogen bond. Fluorine is the most electronegative element in the periodic table. As a result, the F-H-F bond will be the most powerful H bond.

Which of the following is most stable H2? ›

H2+ is most stable as one electron is being shared with two nuclei.

Which bond is stronger H2 or HF? ›

Bond in H2 is strong because bond length is least. Bond in HF is strongest single bond because of high polarity.

Why H2 is stable but He2 is not stable? ›

The two such half field atomic orbitals compbine to from a molecular orbital which contains both these electrons. But helium (Z=2) has already a filled orbital (1s2). Therefore, the atomic orbitals of the helium atoms donot combine. Thus, a molecule of H2 exists while that of He2 does not.

Why is H2 stable than He2? ›

Answer. H2+ will be more stable than He2+ because while they both have same bond order, He2+ has anti bonding electrons which destabilizes the molecule.

Why is H2 stable and He2 unstable? ›

Because of the number of valence electrons. An atom must have an outermost shell that is completely filled with valence electrons to be stable.

What is the energy of H2+ ion? ›

The curve for ψ+ refers to the ground state of the molecule where a minimum energy is found for a nuclear distance of approximately 2ao (i.e. 100pm). Thus, H2+ should exist as a stable molecule. The calculated bonding energy is 1.77 eV. This is a quite satisfying result; from experiments we get 2.77 eV.

What is the charge number of H2? ›

The oxidation number for elemental hydrogen, H2, is 0.

How many ions are in H2? ›

Answer and Explanation: Hydrogen has two ions.

What is the difference between H2 and H+? ›

H+ is used when the system is under acidic conditions. With H2 molecules in the system, then it would not be acidic, due to their neutral charge.

What does H+ mean ion? ›

Hydrogen ion is the nucleus of a hydrogen atom separated from its accompanying electron. The hydrogen nucleus is made up of a particle carrying a unit positive electric charge, called a proton. The isolated hydrogen ion, represented by the symbol H+, is therefore customarily used to represent a proton.

Does H+ mean acid? ›

Acidic solutions

Any solution with more H+ than OH- is said to contain free H+ ions and is acidic. The definition of an acid is: “A substance with free hydrogen ions (H+) in solution.” Acids have a pH < (less than) 7.

What is H+ called? ›

The terms hydrogen ion H+ and proton, p or p+, are used synonymously in chemistry. A hydrogen ion is a positively charged molecule.

Is hydrogen H2 or H? ›

Hydrogen is the chemical element with the symbol H and atomic number 1. Hydrogen is the lightest element. At standard conditions hydrogen is a gas of diatomic molecules having the formula H 2. It is colorless, odorless, tasteless, non-toxic, and highly combustible.

Where is H2 produced? ›

Hydrogen can be produced from diverse, domestic resources. Currently, most hydrogen is produced from fossil fuels, specifically natural gas. Electricity—from the grid or from renewable sources such as wind, solar, geothermal, or biomass—is also currently used to produce hydrogen.

Is H2+ a stable diatomic species? ›

The simplest stable diatomic species is H2^+ which has a minimum number of (electron/proton).

Which species out of H2 H2 and H2+ is paramagnetic? ›

H2+ is paramagnetic due to the presence of one unpaired electron.

What is the magnetic nature of H2? ›

There is no unpaired electrons in H2. So, the H2 molecule is stable and Diamagnetic.

How do you find the bond length? ›

Bond length can be calculated in three easy steps. First, determine the type of covalent bond between the atoms (single, double or triple). Then, using a covalent radii chart, find the atomic radii in these bonds. Finally, add them together and you have the approximate bond length.

Why does H2 have the shortest bond length? ›

The bond length is also known to be directly proportional to the atomic size, which refers to the fact that, the higher the atomic size, the higher will be the bond length, and vice versa. Here, Hydrogen has the lowest atomic size, hence, H−H has the lowest bond length.

What is the bond length of H2 o2? ›

In solid phase O−H bond length in H2O2 is 98.8 pm and in gaseous phase it is 95 pm. But O−H bond length in water is always 96pm.

What kind of bond is H2? ›

Covalent molecules made of only one type of atom, like hydrogen gas (H2), are nonpolar because the hydrogen atoms share their electrons equally.

What is bond length with example? ›

Bond length is a property of a chemical bond between types of atoms. Bonds vary between atoms depending on the molecule that contains them. For example the carbon-hydrogen bond is different in methyl chloride as is methane. When more electrons participate in a bond, it tends to be shorter.

How do you calculate bond order and length? ›

Bond order (B.O.) is defined as one-half the difference between the number of electrons present in the bonding and the anti-bonding orbitals. Bond order (B.O) = ( Number of bonding electrons - Number of the anti-bonding electron) Bond order = (Nb- Na)

What is bond length and size? ›

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. The bigger the size of the atom, the longer the bond length. The bond length decreases in bond multiplicity. It is measured by spectroscopic, X-ray diffraction, rotational spectroscopy, etc.

Why the bond in H2 is longer than H2? ›

The bond length in H2+ is longer than in H2 because only one electron is present to shield the two nuclei from mutual repulsion. In H2,there are two electrons to hold the two nuclei ,thus nuclear repulsion is less than that in H+2. Hence, nuclear separation in H+2 is more than in H2 .

Is bond length of H2+ longer than H2? ›

bond length order is as follows: H2+ is greater than H2- which is greater than H2. Explanation: Bond length is the average distance between the nuclei of the two bonded atom in a molecule.

Why do H2 bonds form? ›

Why Hydrogen Bonds Form. The reason hydrogen bonding occurs is because the electron is not shared evenly between a hydrogen atom and a negatively charged atom. Hydrogen in a bond still only has one electron, while it takes two electrons for a stable electron pair.

Which has shortest bond length H2 or N2? ›

N2 has the shortest bond length as its bond order is largest.

What is the bond in H2 o2? ›

Hydrogen peroxide H2O2 has a structure H—O—O—H, where each dash in this structural formula is a single covalent bond. That is, each bond involves a pair of shared electrons, formally with one electron from each of the atoms at the ends of the bond.

Is h2 an element or ion? ›

h2 is an element but h2o is a compound.


1. Calculating pH, pOH, [H+], [H3O+], [OH-] of Acids and Bases - Practice
2. Calculating pH and H3O+ Contentration
3. How to Balance NH3 + O2 = NO2 + H2O (Ammonia + Oxygen gas)
(Wayne Breslyn)
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