The hybrid states of phosphorous atoms in $P C l_{5}$ and $P B r_{3}$ in gaseous phase are $s p^{3} d$ . But in solid $P C l_{5}$ , phosphorous shows $s p^{3} d^{2}$ hybrid state. While $P$ in $P B r$ is in $s p^{3}$ hybrid state. This is because :

Question

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Answer 1

Both $P C l_{5}$ and $P B r_{5}$ have triagonal bipyramidal geometry this is not a regular structure and is not very stable.

Therefore $P C l_{5}$ splits up into two more stable octahedral and tetrahedral structures .

$P C l_{5}$ in solid form exists as $[P C l_{4}]^{+} [P C l_{6}]^{-}$ . The hybridisation state of $P$ in $[P C l_{4}]^{+}$ and $[P C l_{6}]^{-}$ is
$s p^{3}$ and $s p^{3} d^{2}$ respectively.

$P B r_{5}$ in solid form exists as $[P B r_{4}]^{+} B r^{-}$ and hybridisation state of $P$ in $[P B r_{4}]^{+}$ is $s p^{3}$ hybridised.

Answer 2

Both

P C l_{5}

and

P B r_{5}

have triagonal bipyramidal geometry this is not a regular structure and is not very stable.

Therefore

P C l_{5}

splits up into two more stable octahedral and tetrahedral structures .

P C l_{5}

in solid form exists as

[P C l_{4}]^{+} [P C l_{6}]^{-}

. The hybridisation state of

P

in

[P C l_{4}]^{+}

and

[P C l_{6}]^{-}

is

s p^{3}

and

s p^{3} d^{2}

respectively.

P B r_{5}

in solid form exists as

[P B r_{4}]^{+} B r^{-}

and hybridisation state of

P

in

[P B r_{4}]^{+}

is

s p^{3}

hybridised.

The hybrid states of phosphorous atoms in $P C l_{5}$ and $P B r_{3}$ in gaseous phase are $s p^{3} d$ . But in solid $P C l_{5}$ , phosphorous shows $s p^{3} d^{2}$ hybrid state. While $P$ in $P B r$ is in $s p^{3}$ hybrid state. This is because :

$P C l_{5}$ in solid form exists as $[P C l_{4}]^{+} [P C l_{6}]^{-}$

$P B r_{5}$ in solid form exists as $[P C l_{4}]^{+} [P B r_{6}]^{-}$

$P C l_{5}$ in solid form exists as $[P C l_{4}]^{+} C l^{-}$

$P B r_{5}$ in solid form exists as $[P B r_{4}]^{+} B r^{-}$

In which of the following 'N' atom is not $s p^{2}$ hybridised?

sp-hybridization is involved in the oxides:

Each N-atom is $s p^{3}$ - hybridised in $N_{2} H_{4}$

It has gauche conformation

Read the above assertion and reason and choose the correct statement regarding it.

Each N-atom is $s p^{3}$ - hybridised in $N_{2} H_{4}$

$P C l_{5}$ is a covalent in gaseous and liquid states but ionic in solid state.

$P C l_{5}$ in solid state consists of tetrahedral $P C l_{4}^{+}$ cation and octahedral $P C l_{6}^{-}$ anion.

In solid state $P C l_{5}$ exist as an ionic solid $[P C l_{4}]^{+} [P C l_{6}]^{-}$ hybridisation state of cation and anion respectively :

Revise with Concepts

Compounds of Phosphorus - PCl3

Compounds of Phosphorus - PCl5

The hybrid states of phosphorous atoms in PCl5​ and PBr3​ in gaseous phase are sp3d. But in solid PCl5​, phosphorous shows sp3d2 hybrid state. While P in PBr is in sp3 hybrid state. This is because :

PCl5​ in solid form exists as [PCl4​]+[PCl6​]−

PBr5​ in solid form exists as [PCl4​]+[PBr6​]−

PCl5​ in solid form exists as [PCl4​]+Cl−

PBr5​ in solid form exists as [PBr4​]+Br−

In which of the following 'N' atom is not sp2 hybridised?

sp-hybridization is involved in the oxides:

Each N-atom is sp3 - hybridised in N2​H4​

It has gauche conformation Read the above assertion and reason and choose the correct statement regarding it.

Each N-atom is sp3 - hybridised in N2​H4​

PCl5​ is a covalent in gaseous and liquid states but ionic in solid state.

PCl5​ in solid state consists of tetrahedral PCl4+​ cation and octahedral PCl6−​ anion.

In solid state PCl5​ exist as an ionic solid [PCl4​]+[PCl6​]− hybridisation state of cation and anion respectively :

Revise with Concepts

Compounds of Phosphorus - PCl3

Compounds of Phosphorus - PCl5

The hybrid states of phosphorous atoms in $P C l_{5}$ and $P B r_{3}$ in gaseous phase are $s p^{3} d$ . But in solid $P C l_{5}$ , phosphorous shows $s p^{3} d^{2}$ hybrid state. While $P$ in $P B r$ is in $s p^{3}$ hybrid state. This is because :

$P C l_{5}$ in solid form exists as $[P C l_{4}]^{+} [P C l_{6}]^{-}$

$P B r_{5}$ in solid form exists as $[P C l_{4}]^{+} [P B r_{6}]^{-}$

$P C l_{5}$ in solid form exists as $[P C l_{4}]^{+} C l^{-}$

$P B r_{5}$ in solid form exists as $[P B r_{4}]^{+} B r^{-}$

In which of the following 'N' atom is not $s p^{2}$ hybridised?

Each N-atom is $s p^{3}$ - hybridised in $N_{2} H_{4}$

It has gauche conformation

Read the above assertion and reason and choose the correct statement regarding it.

Each N-atom is $s p^{3}$ - hybridised in $N_{2} H_{4}$

$P C l_{5}$ is a covalent in gaseous and liquid states but ionic in solid state.

$P C l_{5}$ in solid state consists of tetrahedral $P C l_{4}^{+}$ cation and octahedral $P C l_{6}^{-}$ anion.

In solid state $P C l_{5}$ exist as an ionic solid $[P C l_{4}]^{+} [P C l_{6}]^{-}$ hybridisation state of cation and anion respectively :