Ncert Exercises & Solutions

From the perspective of molecular orbital theory, which statement is false?

1. ${Be}_{2}$ is not a stable molecule.

2. ${He}_{2}$ is not stable but ${He}_{2}^{+}$ is expected to exist.

3. Bond strength of $N_{2}$ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

4. The order of energies of molecular orbitals in $N_{2}$ molecule is:

$σ_{2 s} < σ_{2 s}^{*} < σ_{2 p_{z}} < (π_{2 p_{x}} \approx π_{2 p_{y}})$
$< (π_{2 p_{x}}^{*} \approx π_{2 p_{y}}^{*}) < σ_{2 p_{z}}^{*}$

HINT: Bond order =

\frac{1}{2} [N_{b}

-

N_{a}]

Explanation:

Step 1:

Existence of molecule, bonding nature and energy order of molecular orbitals can be explained on the basis of molecular orbital theory as follows:

(i) Molecules having zero bond order never exists while molecular having non-zero bond order is either exists or expected to exist.

(ii) Higher the value of bond order, higher will be its bond strength. Electrons present in bonding molecular orbital are known as bonding electrons (N_b) and electrons present on anti-bonding molecular orbital are known as anti-bonding electrons (N_a) and half of their difference is known as bond order i.e.,

Step 2:

Molecular orbital diagram of given molecule and ion is as follows:

(a)

{Be}_{2}

(4 + 4 = 8) = σ_{1 s^{2}},

σ_{1 s^{2}}^{*},

σ_{2 s^{2}},

σ_{2 s^{2}}^{*}

Bond order (BO)

= \frac{1}{2}

[Number of bonding electrons (N_b)-Number of anti bonding electrons N_a]

= \frac{4 - 4}{2} = 0

Here, bond order of Be₂ is zero. Thus, it does not exist.

(b)

{He}_{2}

(2 + 2 = 4) = σ_{1 s^{2}},

σ_{1 s^{2}}^{*}

= \frac{2 - 2}{2} = 0

Here, bond order of Be₂ is zero. Hence, it does not exist.

{He}_{2}^{+}

(2 + 2 - 1 = 3) = σ_{1 s^{2}},

σ_{1 s^{1}}^{*}

= \frac{2 - 1}{2} = 0.5

Since, the bond order is not zero, this molecule is expected to exist.

(c)

N_{2}

(7 + 7 = 14) = σ_{1 s^{2}},

σ_{1 s^{2}}^{*},

σ_{2 s^{2}},

σ_{2 s^{2}}^{*},

π_{2 {p_{x}}^{2}} \approx π_{2 {p_{y}}^{2}},

σ_{2 {p_{z}}^{2}}

= \frac{10 - 4}{2} = 3

Thus, dinitrogen (N₂) molecule contain triple bond and no any molecule of second period have more than double bond. Hence, bond strength of N₂ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

(d) It is incorrect. The correct order of energies of molecular orbitals in N₂ molecule is

σ_{2 s} < σ_{2 s}^{*} < (π_{2 p_{x}} \approx π_{2 p_{y}}) < σ_{2 p_{z}} < (π_{2 p_{x}}^{*} \approx π_{2 p_{y}}^{*}) < σ_{2 p_{z}}^{*}

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