Q: The d-orbital involved in \$sp^{3}d\$ hybridisation is :

The linear combination of a \$4s-\$orbital with three \$4p-\$orbitals (\$4px\$, \$4py\$, and \$4pz\$) and one \$3d-\$orbital (3\$d_{z^{2}}\$) results in five \$sp^{3}d\$-orbitals

Q: The hybridisation of \$Ni\$ in \$Ni(CO)_4\$ is:

The oxidation state of \$Ni\$ is 0. The valence shell electronic configuration of the ground state of \$Ni\$ is \$3d^84s^2\$, but all the 10 electrons are pushed into \$3d\$ orbital due to the strong field \$CO\$ ligands approaching the \$Ni\$ atom, thereby forming \$sp^3\$ hybridization.

Q: The correct geometry and hybridization for \$XeF_4\$ are :

\$XeF_4\$ is \$AB_4L_2\$ in which the Xenon is bonded with four fluorine atoms with 4 sigma bonds and xenon has two lone pairs (4 sigma bonds and 2 lone pairs).So, the hybridisation is \$sp^{3}d^{2}\$.Geometry is octahedral and shape is square planar.

Q: Which of the following has \$p\pi -d\pi\$ bonding?

In \$SO^{2-}_3\$ ion, the \$sp^3\$ is hybridised, so In S the, three p-orbitals forms a bonds with three oxygen atoms and unhybridised & orbital is involved in it bond formation. \$_{16}O = 1s^2, 2s^2, 2p_x^2, 2p_y^1, 2p_x^1\$In oxygen, two unpaired. p-orbitals are present, one is involved in \$\sigma\$-bond formation while other is used in \$\pi\$-bond formation. Thus, in \$SO^{2-}_3, p_{\pi}\$ and \$p_{\pi}\$ orbitals are involved for \$p_{\pi}-d_{\pi}\$ bonding.

Q: Which hybrid orbitals are used by carbon atoms in the following molecules?\$(a) \displaystyle CH_{3}-CH_{3}\$\$ (b) CH_{3}-CH=CH_{2}\$\$(c) CH_{3}-CH_{2}-OH\$\$(d) CH_{3}-CHO\$ \$(e) CH_{3}COOH \$

\$(a) \displaystyle CH_{3}-CH_{3}\$ Both the carbon atoms are \$\displaystyle sp^3 \$ hybridized.\$(b) \displaystyle CH_{3}-CH=CH_{2} \$ The carbon atom of methyl group is \$\displaystyle sp^3\$ hybridized whereas other two carbon atoms are \$\displaystyle sp^2 \$ hybridized.\$(c) \displaystyle CH_{3}-CH_{2}-OH \$ Both the carbon atoms are \$\displaystyle sp^3 \$ hybridized.\$(d) \displaystyle CH_{3}-CHO \$ The carbon atom of methyl group is \$\displaystyle sp^3 \$ hybridized whereas the other carbon atom is \$\displaystyle sp^2\$ hybridized.\$(e) \displaystyle CH_{3}COOH \$ The carbon atom of methyl group is \$\displaystyle sp^3 \$ hybridized whereas the other carbon atom is \$\displaystyle sp^2 \$ hybridized.

Q: Find out shape and hybridisation of the following molecules ( \$PCl_5\$, \$NH_3\$, \$H_2O\$, \$SF_6\$ )

\$PCl_{5}\$ : \$P(G.S)=1s^{2}\,\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{3}\$\$P(E.S)=1s^{2}\,\,2s^{2}\,2p^{6}\,3s^{1}\,3p^{3}\$One \$s\$ , three \$p\$ & one \$d\$ orbitals of same shell participate to give \$sp^{3}d \$ hybridisation.Shape: Trigonal bipyramidal.Figure-1 represents the shell diagram.Figure-2 represents the shape .\$NH_{3}: N(G.S)=1s^{2}\,\,2s^{2}\,2p^{3}\$\$N(E.S)=1s^{2}\$One \$s\$ , three \$p\$ participate to give \$sp^{3}\$ hybridisation.Shape: PyramidalFigure-3 represents the shell diagram.Figure-4 represents the shape .\$H_{2}O: O(G.S)=1s^{2}\,\,2s^{2}\,2p^{4}\$\$O(E.S)=1s^{2}\,2s^{2}\,2p_{x}^{2}\,2p_{y}^{1}\,2p_{z}^{1}\$Two half filled \$sp^{3}\$ hybrid orbital of Oxygen atom & \$s\$ orbital of 2 Hydrogen atom overlap to form water.Shape: V-shaped or distorted tetrahedral due to \$lp-lp,lp-bp,bp-bp\$ repulsion.Figure-5 represents the shell diagram.Figure-6 represents the shape .\$SF_{6}: 1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}3p^{4}\$Hybridisation- \$sp^{3}d^{2}\$ Shape: Octahedral.Figure-7 represents the shell diagram.Figure-8 represents the shape .

Q: The hybridisation of carbon atoms in \$C-C\$ single bond of \$HC \equiv C - CH = CH_2\$ is:

The hybridisation of carbon atoms in \$\displaystyle C−C\$ single bond of \$\displaystyle HC \equiv C - CH = CH_2\$ is \$\displaystyle sp - sp^2\$. One \$\displaystyle C\$ atom is bonded to another \$\displaystyle C\$ atom by a triple bond and is \$\displaystyle sp\$ hybridised. Other \$\displaystyle C\$ atom is attached to another \$\displaystyle C\$ by a double bond and is \$\displaystyle sp^2\$ hybridised.

Question 1

What are hybridisation states of each carbon atom in the following compounds?$CH_2=C=O,\, CH_3CH=CH_2,\, (CH_3)_2CO,\, CH_2=CHCN,\, C_6H_6$

Accepted Answer

The hybridisation states of each carbon atom in the following compounds are given below. $\underset {sp^2}{C}H_2=\underset {sp}{C}=O,\, \underset {sp^3}{C}H_3\underset {sp^2}{C}H=\underset {sp^2}{C}H_2,\, (\underset {sp^3}{C}H_3)_2\underset {sp^2}{C}O,\, \underset {sp^2}{C}H_2=\underset {sp^2}{C}H\underset {sp^3}{C}N,\, \underset {sp^2}{C_6}H_6$

Question 2

The hybridization of atomic orbitals of nitrogen in $NO_2^{+}$, $NO_3^{-}$ and $NH_4^{+}$ are:

Accepted Answer

Here's a nice short trick.1) Count and add all the electrons in valence shell2) Divide it by $8$3) On dividing by $8$ you will get a remainder and quotient, add quotient $+$ (remainder / $2$)4) Now get the hybridization corresponding to the number what you gotIf $2$ its $sp$,if $3$ its $sp^{2}$, if  $4$ its $sp^{3}$, if $5$ its $sp^{3}d$ and so on.Ex: $NO^{2+}$-> Total electrons in valence shell : $5 + 6 \times 2 - 1 = 16$.->$ 16/8 = 2$. Quotient = $2$ and Remainder =$ 0$-> $2 + 0/2 = 2$ i.e $sp$$NH^{4+}$-> Total electrons in valence shell : $5 + 7 \times 4 - 1 = 32$.-> $32/8 = 4$. Quotient = $4$ and Remainder = $0$-> $4 + 0/2 = 4$ i.e $sp^{3}$$NO^{3-}$-> Total electrons in valence shell : $5 + 6 \times 3 +1 = 24$.-> $24/8 = 3$. Quotient = $3$ and Remainder = $0$-> $3 + 0/2 = 3$ i.e $sp^{2}$Hence option B is correct.

Question 3

The hybridisation in $P{Cl}_{5}$ molecule is:

Accepted Answer

Phosphorus has 5 valence electrons. 1 electron each is shared by chlorine atom hence, $PCl_5$ molecule has $sp^3d$ hybridisation and hence trigonal bipyramidal shape.

Question 4

The d-orbital involved in $sp^{3}d$ hybridisation is :

Accepted Answer

The linear combination of a $4s-$orbital with three $4p-$orbitals ($4px$, $4py$, and $4pz$) and one $3d-$orbital (3$d_{z^{2}}$) results in five $sp^{3}d$-orbitals

Question 5

The hybridisation of $Ni$ in $Ni(CO)_4$ is:

Accepted Answer

The oxidation state of $Ni$ is 0. The valence shell electronic configuration of the ground state of $Ni$ is $3d^84s^2$, but all the 10 electrons are pushed into $3d$ orbital due to the strong field $CO$ ligands approaching the $Ni$ atom, thereby forming $sp^3$ hybridization.

Question 6

The correct geometry and hybridization for $XeF_4$ are :

Accepted Answer

$XeF_4$ is $AB_4L_2$ in which the Xenon is bonded with four fluorine atoms with 4 sigma bonds and xenon has two lone pairs (4 sigma bonds and  2 lone pairs).So, the hybridisation is $sp^{3}d^{2}$.Geometry is octahedral and shape is square planar.

Question 7

Which of the following has $p\pi -d\pi$ bonding?

Accepted Answer

In $SO^{2-}_3$ ion, the $sp^3$ is hybridised, so In S the, three p-orbitals forms a bonds with three oxygen atoms and unhybridised & orbital is involved in it bond formation. $_{16}O = 1s^2, 2s^2, 2p_x^2, 2p_y^1, 2p_x^1$In oxygen, two unpaired. p-orbitals are present, one is involved in $\sigma$-bond formation while other is used in $\pi$-bond formation. Thus, in $SO^{2-}_3, p_{\pi}$ and $p_{\pi}$ orbitals are involved for $p_{\pi}-d_{\pi}$ bonding.

Question 8

Which hybrid orbitals are used by carbon atoms in the following molecules?$(a) \displaystyle CH_{3}-CH_{3}$$ (b)  CH_{3}-CH=CH_{2}$$(c) CH_{3}-CH_{2}-OH$$(d)  CH_{3}-CHO$ $(e)  CH_{3}COOH $

Accepted Answer

$(a) \displaystyle CH_{3}-CH_{3}$ Both the carbon atoms are $\displaystyle sp^3 $ hybridized.$(b) \displaystyle CH_{3}-CH=CH_{2} $ The carbon atom of methyl group is $\displaystyle sp^3$ hybridized whereas other two carbon atoms are $\displaystyle sp^2 $ hybridized.$(c) \displaystyle CH_{3}-CH_{2}-OH $ Both the carbon atoms are $\displaystyle sp^3 $ hybridized.$(d) \displaystyle CH_{3}-CHO $ The carbon atom of methyl group is $\displaystyle sp^3 $ hybridized whereas the other carbon atom is $\displaystyle sp^2$ hybridized.$(e) \displaystyle CH_{3}COOH  $ The carbon atom of methyl group is $\displaystyle sp^3 $ hybridized whereas the other carbon atom is $\displaystyle sp^2 $ hybridized.

Question 9

Find out shape and hybridisation of the following molecules ( $PCl_5$, $NH_3$, $H_2O$, $SF_6$ )

Accepted Answer

$PCl_{5}$ : $P(G.S)=1s^{2}\,\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{3}$$P(E.S)=1s^{2}\,\,2s^{2}\,2p^{6}\,3s^{1}\,3p^{3}$One $s$ , three $p$ & one $d$ orbitals of same shell participate to give $sp^{3}d $ hybridisation.Shape: Trigonal bipyramidal.Figure-1 represents the shell diagram.Figure-2 represents the shape .$NH_{3}: N(G.S)=1s^{2}\,\,2s^{2}\,2p^{3}$$N(E.S)=1s^{2}$One $s$ , three $p$ participate to give $sp^{3}$ hybridisation.Shape: PyramidalFigure-3 represents the shell diagram.Figure-4 represents the shape .$H_{2}O: O(G.S)=1s^{2}\,\,2s^{2}\,2p^{4}$$O(E.S)=1s^{2}\,2s^{2}\,2p_{x}^{2}\,2p_{y}^{1}\,2p_{z}^{1}$Two half filled $sp^{3}$ hybrid orbital of Oxygen atom & $s$ orbital of 2 Hydrogen atom overlap to form water.Shape: V-shaped or distorted tetrahedral due to $lp-lp,lp-bp,bp-bp$ repulsion.Figure-5 represents the shell diagram.Figure-6 represents the shape .$SF_{6}: 1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}3p^{4}$Hybridisation- $sp^{3}d^{2}$ Shape: Octahedral.Figure-7 represents the shell diagram.Figure-8 represents the shape .

Question 10

The hybridisation of carbon atoms in $C-C$ single bond of $HC \equiv C - CH = CH_2$ is:

Accepted Answer

The hybridisation of carbon atoms in $\displaystyle C&#8722;C$ single bond of $\displaystyle HC \equiv C - CH = CH_2$ is $\displaystyle sp - sp^2$. One $\displaystyle C$ atom is bonded to another $\displaystyle C$ atom by a triple bond and is $\displaystyle sp$ hybridised. Other $\displaystyle C$ atom is attached to another $\displaystyle C$ by a double bond and is $\displaystyle sp^2$ hybridised.

Revise with Concepts

Hybridisation - Introduction

Types of Hybridisation

Quick Summary With Stories

Hybridization

Types of Hybridisation -sp Hybridisation

Types of Hybridisation -sp2 Hybridisation

Types of Hybridisation -SP3 Hybridisation

Hybridisation of Elements Involving d-orbitals

What are hybridisation states of each carbon atom in the following compounds?
$C H_{2} = C = O, C H_{3} C H = C H_{2}, (C H_{3})_{2} C O, C H_{2} = C H C N, C_{6} H_{6}$

The hybridization of atomic orbitals of nitrogen in $N O_{2}^{+}$ , $N O_{3}^{-}$ and $N H_{4}^{+}$ are:

The hybridisation in $P C l_{5}$ molecule is:

The d-orbital involved in $s p^{3} d$ hybridisation is :

The hybridisation of $N i$ in $N i (C O)_{4}$ is:

The correct geometry and hybridization for $X e F_{4}$ are :

Which of the following has $p π - d π$ bonding?

Which hybrid orbitals are used by carbon atoms in the following molecules?
$(a) C H_{3} - C H_{3}$
$(b) C H_{3} - C H = C H_{2}$
$(c) C H_{3} - C H_{2} - O H$
$(d) C H_{3} - C H O$
$(e) C H_{3} C O O H$

Find out shape and hybridisation of the following molecules ( $P C l_{5}$ , $N H_{3}$ , $H_{2} O$ , $S F_{6}$ )

The hybridisation of carbon atoms in $C - C$ single bond of $H C \equiv C - C H = C H_{2}$ is:

Revise with Concepts

Hybridisation - Introduction

Types of Hybridisation

Quick Summary With Stories

Hybridization

Types of Hybridisation -sp Hybridisation

Types of Hybridisation -sp2 Hybridisation

Types of Hybridisation -SP3 Hybridisation

Hybridisation of Elements Involving d-orbitals

What are hybridisation states of each carbon atom in the following compounds? CH2​=C=O,CH3​CH=CH2​,(CH3​)2​CO,CH2​=CHCN,C6​H6​

The hybridization of atomic orbitals of nitrogen in NO2+​, NO3−​ and NH4+​ are:

The hybridisation in PCl5​ molecule is:

The d-orbital involved in sp3d hybridisation is :

The hybridisation of Ni in Ni(CO)4​ is:

The correct geometry and hybridization for XeF4​ are :

Which of the following has pπ−dπ bonding?

Which hybrid orbitals are used by carbon atoms in the following molecules?(a)CH3​−CH3​(b)CH3​−CH=CH2​(c)CH3​−CH2​−OH(d)CH3​−CHO (e)CH3​COOH

Find out shape and hybridisation of the following molecules ( PCl5​, NH3​, H2​O, SF6​ )

The hybridisation of carbon atoms in C−C single bond of HC≡C−CH=CH2​ is: