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(C) Hybridization is a concept used to explain the geometry of molecules by mixing atomic orbitals. $(i)$ $BCl_3$: Boron $(Z=5)$ has ground state conf

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$(ii), (iii)$

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(C) Hybridization is a concept used to explain the geometry of molecules by mixing atomic orbitals. $(i)$ $BCl_3$: Boron $(Z=5)$ has ground state configuration $1s^2 2s^2 2p^1$. To form $3$ bonds,it promotes an electron to $2p$ to form $sp^2$ hybrid orbitals. $(ii)$ $NH_3$: Nitrogen $(Z=7)$ has $1s^2 2s^2 2p^3$. It uses $sp^3$ hybridization in the ground state to form $3$ bonds and hold a lone pair. $(iii)$ $PCl_3$: Phosphorus $(Z=15)$ has $3s^2 3p^3$. It uses $sp^3$ hybridization in the ground state to form $3$ bonds and hold a lone pair. $(iv)$ $BeF_2$: Beryllium $(Z=4)$ has $1s^2 2s^2$. It promotes an electron to $2p$ to form $sp$ hybrid orbitals. Therefore,$(ii)$ and $(iii)$ undergo hybridization in their ground state electronic configurations without needing promotion of electrons to higher energy subshells.

$A. NO_3^-$	$1. sp^3$
$B. NH_2^-$	$2. sp$
$C. SCN^-$	$3. sp^3d$
$D. ICl_2^-$	$4. sp^2$

Choose the molecules in which hybridization occurs in the ground state?
$(i)$ $BCl_3$ $(ii)$ $NH_3$
$(iii)$ $PCl_3$ $(iv)$ $BeF_2$
The correct answer is

Similar Questions

Total number of species from the following with central atom utilising $sp^2$ hybrid orbitals for bonding is.................... $NH_3, SO_2, SiO_2, BeCl_2, C_2H_2, C_2H_4, BCl_3, HCHO, C_6H_6, BF_3, C_2H_4Cl_2$

$CCl_4$ has the hybridization:

Match the following ions with the corresponding hybridisation of their central atoms.
$A. NO_3^-$ $1. sp^3$
$B. NH_2^-$ $2. sp$
$C. SCN^-$ $3. sp^3d$
$D. ICl_2^-$ $4. sp^2$

Non-axial $d-$ orbital is used in hybridisation of

$X$ and $Y$ are two covalent molecules in which the hybridization of the central atoms is the same,but their shapes are different. $X$ and $Y$ are:

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Choose the molecules in which hybridization occurs in the ground state?$(i)$ $BCl_3$ $(ii)$ $NH_3$$(iii)$ $PCl_3$ $(iv)$ $BeF_2$The correct answer is