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Q1:

Select the correct statement regarding $X e O_{2} F_{2}$ among the folowing.

a. Xe is $s p^{3} d$ hybridised.

b. 4 bond pairs and 1 lone pair are present around Xe.

c. Shape of $X e O_{2} F_{2}$ is square pyramidal.

d. Contains $dπ - p π$ bonds.

1. a and b

2. a, and d

3. a, b and d

4. b, c and d

Subtopic: Hybridisation |

58%

Level 3: 35%-60%

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Q2:

Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.

Column I	Column II
A. XeF₆	1. Distorted octahedral
B. XeO₃	2. Square planar
C. XeOF₄	3. Pyramidal
D. XeF₄	4. Square pyramidal

	A	B	C	D
1.	1	2	4	3
2.	4	3	1	2
3.	4	1	2	3
4.	1	3	4	2

Subtopic: Hybridisation |

80%

Level 1: 80%+

NEET - 2016

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Q3:

Match the column :

	I		II
(P)	${BH}_{4}^{-}$	(I)	Linear; 2 bond pair and 3 lone pair
(Q)	${ICl}_{2}^{+}$	(II)	Tetrahedral, 4 bond pair and no lone pair
(R)	${ICl}_{2}^{-}$	(III)	Tetrahedral; 3 bond pair and 1 lone pair
(S)	${ICl}_{4}^{-}$	(IV)	V shaped; 2 bond pair and 2 lone pair
		(V)	Square planar; 4 bond pair and 2 lone pair
		(VI)	Square planar; 4 bond pair and no lone pair

1. P = II, Q = IV, R = III, S = VI

2. P = II, Q = IV, R = I, S = V

3. P = II, Q = I, R = V, S = VI

4. P = II, Q = I, R = III, S = IV

Subtopic: Hybridisation | V.S.E.P.R & V.B.T |

84%

Level 1: 80%+

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Q4:

Match the compounds of Xe in column I with the molecular structure in column II.

Column-I	Column-II
(a) ${XeF}_{2}$	(i) Square planar
(b) ${XeF}_{4}$	(ii) Linear
(c) ${XeO}_{3}$	(iii) Square pyramidal
(d) ${XeOF}_{4}$	(iv) Pyramidal

	(a)	(b)	(c)	(d)
1.	(ii)	(i)	(iii)	(iv)
2.	(ii)	(iv)	(iii)	(i)
3.	(ii)	(iii)	(i)	(iv)
4.	(ii)	(i)	(iv)	(iii)

Subtopic: Hybridisation | V.S.E.P.R & V.B.T |

80%

Level 1: 80%+

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Q5:

Match the species in Column I with the type of hybrid orbitals in Column II.

	Column I		Column II
A.	SF₄	1.	sp³d²
B.	IF₅	2.	d²sp³
C.	NO₂⁺	3.	sp³d
D.	NH₄⁺	4.	sp³
		5.	sp

Codes:

Options:	A	B	C	D
1.	3	1	5	4
2.	1	2	3	5
3.	5	4	3	2
4.	4	5	3	2

Subtopic: Hybridisation | V.S.E.P.R & V.B.T |

90%

Level 1: 80%+

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Q6:

Match the species in Column I with the shape in Column II.

	Column I		Column II
A.	H₃O⁺	1.	Linear
B.	HC $\equiv$ CH	2.	Angular
C.	ClO₂^-	3.	Tetrahedral
D.	NH₄⁺	4.	Trigonal bipyramidal
		5.	Pyramidal

Codes:

Options:	A	B	C	D
1.	5	1	2	3
2.	1	2	3	5
3.	5	4	3	2
4.	4	5	3	2

Subtopic: Hybridisation | V.S.E.P.R & V.B.T |

89%

Level 1: 80%+

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Q7:

Match the shape of molecules in Column I with the type of hybridization in Column II.

	Column I		Column II
A.	Tetrahedral	1.	sp²
B.	Trigonal	2.	sp
C.	Linear	3.	sp³

Codes:

Options:	A	B	C
1.	3	1	2
2.	1	2	3
3.	5	4	3
4.	4	5	3

Subtopic: Hybridisation |

95%

Level 1: 80%+

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Q8:

Match the coordination number and type of hybridization with the distribution of hybrid orbitals in space based on Valence bond theory.

	Coordination number and type of hybridisation		Distribution of hybrid orbitals in space
(a)	4, sp³	(i)	Trigonal bipyramidal
(b)	4, dsp²	(ii)	Octahedral
(c)	5, sp³d	(iii)	Tetrahedral
(d)	6, d²sp³	(iv)	Square planar

	(a)	(b)	(c)	(d)
1.	(ii)	(iii)	(iv)	(i)
2.	(iii)	(iv)	(i)	(ii)
3.	(iv)	(i)	(ii)	(iii)
4.	(iii)	(i)	(iv)	(ii)

Subtopic: Hybridisation |

85%

Level 1: 80%+

NEET - 2020

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Q9:

Assertion (A):	On heating PCl₅, it readily converts into PCl₃ and Cl₂
Reason (R):	The axial P-Cl bond is larger than the equitorial P-Cl bond

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	Both (A) and (R) are False.

Subtopic: Hybridisation |

Level 3: 35%-60%

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Q10:

Assertion (A):	C₃O₂ is a linear structure.
Reason (R):	Each C atom in C₃O₂ is sp hybridized.

1.	Both (A) and (R) are true and (R) is the correct explanation of (A).
2.	Both (A) and (R) are true but (R) is not the correct explanation of (A).
3.	(A) is true but (R) is false.
4.	Both (A) and (R) are false.

Subtopic: Hybridisation |

76%

Level 2: 60%+

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