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Match the compounds given in Column I wi...

Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option.
`{:("Column I","Column II"),((A) XeF_(6),(1)sp^(3)d^(3)-"distorted octahedral"),((B)XeO_(3),(2)sp^(3)d^(2)-"square planar"),((C)XeOF_(4),(3)sp^(3)-"pyramidal"),((D)XeF_(4),(4)sp^(3)d^(2)-"square pyramidal"):}`

A

`{:(A(1),B(3),C(4),D(2)):}`

B

`{:(A(1),B(2),C(4),D(3)):}`

C

`{:(A(4),B(3),C(1),D(2)):}`

D

`{:(A(4),B(1),C(2),D(3)):}`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question of matching the compounds in Column I with their respective hybridization and shape in Column II, we will analyze each compound step by step. ### Step 1: Analyze XeF₆ - **Valence Electrons**: Xenon (Xe) has 8 valence electrons, and each fluorine (F) contributes 1 electron. Therefore, XeF₆ has a total of 8 + 6 = 14 valence electrons. - **Hybridization**: To accommodate 6 bonding pairs (one for each F), we need to hybridize 6 orbitals. The hybridization is sp³d³. - **Shape**: With 6 bond pairs and 1 lone pair, the geometry is distorted octahedral. **Match**: (A) XeF₆ → (1) sp³d³ - "distorted octahedral" ### Step 2: Analyze XeO₃ - **Valence Electrons**: Xenon has 8 valence electrons, and each oxygen (O) contributes 6 electrons. Therefore, XeO₃ has a total of 8 + 3*6 = 26 valence electrons. - **Hybridization**: XeO₃ has 3 bond pairs and 1 lone pair, leading to sp³ hybridization. - **Shape**: The presence of 3 bond pairs and 1 lone pair gives it a pyramidal shape. **Match**: (B) XeO₃ → (3) sp³ - "pyramidal" ### Step 3: Analyze XeOF₄ - **Valence Electrons**: For XeOF₄, we have 8 (Xe) + 6 (O) + 4 (F) = 18 valence electrons. - **Hybridization**: There are 4 bond pairs and 1 lone pair, leading to sp³d² hybridization. - **Shape**: With 4 bond pairs and 1 lone pair, the shape is square pyramidal. **Match**: (C) XeOF₄ → (4) sp³d² - "square pyramidal" ### Step 4: Analyze XeF₄ - **Valence Electrons**: For XeF₄, we have 8 (Xe) + 4 (F) = 12 valence electrons. - **Hybridization**: There are 4 bond pairs and 2 lone pairs, leading to sp³d² hybridization. - **Shape**: With 4 bond pairs and 2 lone pairs, the shape is square planar. **Match**: (D) XeF₄ → (2) sp³d² - "square planar" ### Final Matching: - (A) XeF₆ → (1) sp³d³ - "distorted octahedral" - (B) XeO₃ → (3) sp³ - "pyramidal" - (C) XeOF₄ → (4) sp³d² - "square pyramidal" - (D) XeF₄ → (2) sp³d² - "square planar" ### Summary of Matches: - A - 1 - B - 3 - C - 4 - D - 2
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Match the compounds give in Column I with the hybridisation and shape given in Column II and mark the correct option.

Sp^3d Hybridisation

Knowledge Check

  • Match the compounds give in Column I with the hybridisation an shape given in Column II and mark the correct option. Codes

    A
    `{:(A,B,C,D),(1,3,4,2):}`
    B
    `{:(A,B,C,D),(1,2,4,3):}`
    C
    `{:(A,B,C,D),(4,3,1,2):}`
    D
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  • Match the compounds given in column I with the hybridisation and shape given in column II and mark the correct option. {:(,"Column I",,"Column II"),((A),XeF_(6),(i),"Distorted octahedral"),((B),XeO_(3),(ii),"Square planar"),((C),XeOF_(4),(iii),"Pyramidal"),((D),XeF_(4),(iv),"Square pyramidal"):}

    A
    `{:(,A,B,C,D),((a),(iv),(iii),(i),"(ii)"):}`
    B
    `{:(,A,B,C,D),((a),(iv),(i),(ii),"(iii)"):}`
    C
    `{:(,A,B,C,D),((a),(i),(iii),(iv),"(ii)"):}`
    D
    `{:(,A,B,C,D),((a),(i),(ii),(iv),"(iii)"):}`

  • Match the species given in Column I with the shape given in Column II and mark the correct option. {:("Column I","Column II"),((A)SF_(4),(1)"Tetrahedral"),((B)BrF_(3),(2)"Pyramidal"),((C)BrO_(3)^(-),(3)"See-saw shaped"),((D)NH_(4)^(+),(4)"Bent T-shaped"):}

    A
    `{:(A(3),B(2),C(1),D(4)):}`
    B
    `{:(A(3),B(4),C(2),D(1)):}`
    C
    `{:(A(1),B(2),C(3),D(4)):}`
    D
    `{:(A(1),B(4),C(3),D(2)):}`

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    Match Column-I with Column-II {:("Column-I","Column-II"),((A)XeF_(2),(p)"Square pyramidal"),((B)PCl_(5),(q)"linear shape"),((C )XeOF_(4),(r )sp^(3)d "hybridization"),((D)XeF_(4), (s) sp^(3)d^(2)):}

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