An atom in a molecule has electrons in 1s, 2s | vedclass

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(C) The atom has electrons in the $3d$ and $4s$ orbitals,meaning it belongs to the $3d$ transition series. Because it possesses $3d, 4s,$ and $4p$ orb

Vedclass · Accepted Answer

$sp^3, dsp^2, d^2sp^3$

Vedclass · Answer

(C) The atom has electrons in the $3d$ and $4s$ orbitals,meaning it belongs to the $3d$ transition series. Because it possesses $3d, 4s,$ and $4p$ orbitals,it can participate in various hybridizations. $sp^3$ hybridization involves $4s$ and $4p$ orbitals (e.g.,$[NiCl_4]^{2-}$). $dsp^2$ hybridization involves one $3d$,one $4s$,and two $4p$ orbitals (e.g.,$[Ni(CN)_4]^{2-}$). $d^2sp^3$ hybridization involves two $3d$,one $4s$,and three $4p$ orbitals (e.g.,$[Cr(NH_3)_6]^{3+}$). Therefore,the atom can undergo $sp^3, dsp^2,$ and $d^2sp^3$ hybridizations.

List-$I$	List-$II$
$(1)$ $SF_4$	$(A)$ $sp^3d^2$
$(2)$ $IF_5$	$(B)$ $d^2sp^3$
$(3)$ $NO_2^+$	$(C)$ $sp^3d$
$(4)$ $NH_4^+$	$(D)$ $sp^3$
	$(E)$ $sp$

An atom in a molecule has electrons in $1s, 2s, 2p, 3s, 3p, 3d$ and $4s$ orbitals. This atom can undergo hybridisations of type

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Match the species in List-$I$ with their correct hybrid orbitals in List-$II$.

List-$I$ List-$II$

$(1)$ $SF_4$ $(A)$ $sp^3d^2$

$(2)$ $IF_5$ $(B)$ $d^2sp^3$

$(3)$ $NO_2^+$ $(C)$ $sp^3d$

$(4)$ $NH_4^+$ $(D)$ $sp^3$

$(E)$ $sp$

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