The incorrect statement is :-

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(B) $1$. The $\pi^* p$ antibonding molecular orbital has two nodal planes.$2$. For $HeH^{+}$,the total number of electrons is $2 + 1 - 1 = 2$. The ele

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Bond order of $HeH^{+}$ is $0.5$

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(B) $1$. The $\pi^* p$ antibonding molecular orbital has two nodal planes.$2$. For $HeH^{+}$,the total number of electrons is $2 + 1 - 1 = 2$. The electronic configuration is $\sigma 1s^2$. Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(2 - 0) = 1$. Thus,the statement that the bond order is $0.5$ is incorrect.$3$. In $NCO^{-}$,the structure is $[	ext{N} \equiv 	ext{C} - 	ext{O}]^-$. The carbon atom is bonded to two atoms with no lone pairs,so it is $sp$ hybridized.$4$. $O_3$ has a bent geometry and is polar,while $O_2$ is a homonuclear diatomic molecule and is non-polar.

List-$I$	List-$II$
$(P). \ IF_7$	$(a). \ \text{Planar} \ \& \ \text{Polar}$
$(Q). \ SO_2$	$(b). \ \text{Non-planar} \ \& \ \text{Polar}$
$(R). \ SF_4$	$(c). \ \text{Planar} \ \& \ \text{Non-polar}$
$(S). \ CS_2$	$(d). \ \text{Non-planar} \ \& \ \text{Non-polar}$

Similar Questions

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$(P). \ IF_7$ $(a). \ \text{Planar} \ \& \ \text{Polar}$

$(Q). \ SO_2$ $(b). \ \text{Non-planar} \ \& \ \text{Polar}$

$(R). \ SF_4$ $(c). \ \text{Planar} \ \& \ \text{Non-polar}$

$(S). \ CS_2$ $(d). \ \text{Non-planar} \ \& \ \text{Non-polar}$

$P_4O_{10}$ has short and long $P-O$ bonds. The number of short $(P=O)$ bonds in this compound is:

What is the total number of planes in $CH_4$ that contain the maximum number of atoms?

Select the correct order of bond angle.

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