Question

Which of the following orders are correct against the property given?
I) dipole moment : $N{F}_3>N{H}_3>B{F}_3$
II) Covalent bond length : $C-O>N-O>O-H$
III) Bond order : $C_2>B_2>H{e}_2$

• A. I, II only
• B. II, III only
• C. I, III only
• D. I, II , III

Answer 1

Answer: (B)

(I) In $B{F}_3$, electronegativity of $F$ -atom is greater than $B$.



In $N{H}_3$, electronegativity of $N$ -atom is greater than $H$. Therefore, direction of all dipole moment with lone pairs lies in same direction, hence net dipole moment increases.



In $N{F}_3$, electronegativity of $F$ is greater than $N$. Therefore, direction of dipole moment are opposite from the lone pair, hence, net dipole decreases.



Correct order of dipole moment will be

$N{H}_3>N{F}_3>B{F}_3$

(II) Covalent bond length

$C-O>N-O>O-H$

as electronegativity difference increases then bond length decreases but lone pair-lone pair repulsion increases the bond length.

(III ) Bond order = Number of bonding electron $\left(N_b\right)$

$\frac{\text{ - Number of antibonding electron }}{2}$

$C_2=\sigma l{s}^2<{\sigma }^{\ast }1s^2<\sigma 2s^2<{\sigma }^{\ast }2s^2<\pi 2p_x^2\simeq \pi 2p_y^2$

$BO=\frac{8-4}{2}=2$

$B_2=\sigma l{s}^2<{\sigma }^{\ast }1s^2<\sigma 2s^2<{\sigma }^{\ast }2s^2<\pi 2p_x^1\simeq \pi 2p_y^1$

$BO=\frac{6-4}{2}=1$

$H{e}_2=\sigma l{s}^2<{\sigma }^{\ast }l{s}^2$

$BO=\frac{2-2}{2}=0$ (does not exist)

II and III statement are correct.