Question

Which of the following orders are correct against the property given?
I) dipole moment : $NF_3 > NH_3 >BF_3$
II) Covalent bond length : $C - O > N - O > O - H$
III) Bond order : $C_2 > B_2 > He_2$

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

Answer 1

Answer: (B)

(I) In $BF _{3}$, electronegativity of $F$ -atom is greater than $B$.



In $NH _{3}$, 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 $NF _{3}$, 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

$NH _{3}>\, NF _{3}>\, BF _{3}$

(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 $\left(N_{b}\right)$

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

$C _{2}=\sigma l s^{2}<\,\sigma^{*} 1 s^{2}<\,\sigma 2 s^{2}<\,\sigma^{*} 2 s^{2}<\pi 2 p_{x}^{2} \simeq \pi 2 p_{y}^{2}$

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

$B _{2}=\sigma l s^{2}<\,\sigma^{*} 1 s^{2}<\,\sigma 2 s^{2}<\,\sigma^{*} 2 s^{2}<\pi 2 p_{x}^{1} \simeq \pi \,2 p_{y}^{1} $

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

$He _{2}=\sigma l s^{2}<\,\sigma^{*} ls ^{2}$

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

II and III statement are correct.