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$S b F_{5}$ reacts with $X e F_{4}$ and $X e F_{6}$ to form ionic compounds $[X e {F_{3}}^{+}]$ $[S b {F_{6}}^{-}]$ and $[X e {F_{5}}^{+}]$ $[S b {F_{6}}^{-}]$ then molecular shape of $[X e {F_{3}}^{+}]$ ion and $[X e {F_{5}}^{+}]$ ion respectively :

Square pyramidal, T-shaped

Bent T-shape, Square pyramidal

See- saw , square pyramidal

Square pyramidal , See-saw

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Explanation

In the reaction between SbF5 and XeF4/XeF6, the ionic compounds formed are [XeF3]+ [SbF6]- and [XeF5]+ [SbF6]-. Based on the VSEPR theory, the molecular shape of [XeF3]+ is bent T-shaped and that of [XeF5]+ is square pyramidal.

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In which of the following species maximum atom can lie in same plane?

$X e F_{2} o_{2}$

$P C l_{5}$

$A s {H_{4}}^{+}$

$X e F_{4}$

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The geometry of ammonia molecule can be best described as:

Nitrogen at one vertex of a regular tetrahedron, the other three vertices being occupied by three hydrogens

Nitrogen at the centre of the tetrahedron, three vertics being occupied by three hydrogens

Nitrogen at the centre of an equilateral triangle, three corners being occupied by three hydrogens

Nitrogen at the junction of a T, three open ends being occupied by three hydrogens

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Explanation

(b) In $\overset{. .}{N} H_{3}$ central atom is nitrogen which is $s p^{3}$ hybrdized hence, it will be at the centre of tetrahedron with H-atom at three vertices.

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Assuming pure 2s and 2p orbitals are carbon are used in forming $C H_{4}$ molecule, which of the following statements is false?

Three C $—$ H bonds will be at right angle

One C $—$ H bond will be weaker then other three C $—$ H bonds

The shape of molecule will be tetrahedral

The angle of C $—$ H bond formed by s-s overlapping will be uncertain with respect to other three bonds

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Which molecular geometry is least likely to result from a trigonal bipyramidal electron geometry?

Trigonal planar

See-saw

Linear

T-shaped

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Explanation

(a) In trigonal by pyramidal geometry lone pair electrons cannot occupy axial positions.

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Which of the following orbital can not form $π$ as well as $δ$ -bond?

$d_{x y}$

$d_{z^{2}}$

$d_{x^{2} - y^{2}}$

$d_{y z}$

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Explanation

(b) $d_{z^{2}}$ can not form $π$ as well as $δ$ bond but it can form $σ$ -bond.

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Give the correct order of initials T and F for following statements. use T if statement is true and F if it is false :

(a) The order of repulsion betweeen different pair of electrons is $l_{p} - l_{p} > l_{p} - b_{p} > b_{p} - b_{p}$

(b) In general , as the number of lone pair of electrons on central atom increases, value of bond angle from normal bond angle also increases

(d) The structure of xeon fluorides and xenon oxyfluorides could not be explained on the basis of VSEPR

TTTF

TFTF

TFTT

TFFF

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Explanation

(b) (ii) (F) In general as the number of lone pair electrons on central atom increases , value of bond angle from normal bond angle decreases due to $l_{p} - l_{p} > l_{p} - b_{p}$

(iv) (F) Structures of xeon fluorides and xeon oxyfluride are explained on the basis VSEPR theory.

In SO $B r_{2,}$ S-O bond has maximum bond length in comparison to S-O bond lengths in $S O F_{2}$ and $S O C l_{2},$ because in $S O B r_{2}$ , S-O bond has been formed by hybrid orbital containing less s- character .

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Which species is planar?

$C {O_{3}}^{2 -}$

$S {O_{3}}^{2 -}$

$C l {O_{3}}^{-}$

$B {F_{4}}^{-}$

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Incorrect statement is:

$A l F_{3} > M g O > M g F_{2}$ : Lattice energy

$L i > N a > A l > M g$ : Electon affinity

$S F_{6} > P F_{5} > S i F_{4}$ : Lewis acidic character

$S I C I_{4} > S i B r_{4} > S i l_{4}$ : Decreasing ionic character

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Explanation

(c) Correct order of lewis acidic character: $P F_{5} > S i F_{4} > S F_{6}$ Although, S has vaccant 3d-orbitals but it cannot accept co-ordinate bond from Lewis base due to steric crowding factor as S-atom is already bonded to six F-atoms.

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What is the geometry of the $I B {r_{2}}^{-}$ ion ?

Linear

Bent shape with bond angle of about $90 °$

Bent shape with bond angle of about $109 °$

Bent shape with bond angle of about $120 °$

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