# NEET Questions Solved

NEET - 2016

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. XeF6        1. Distorted octahedral
B. XeO      2. Square planar
C. XeOF    3. Pyramidal
D. XeF4       4. Square pyramidal

Codes
A B C D
(a) 1 2 4 3
(b) 4 3 1 2
(c) 4 1 2 3
(d) 1 3 4 2

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(d) A-1,B-3,C-4,D-2
The structure of the xenon compunds are represented below :-

Difficulty Level:

• 13%
• 23%
• 10%
• 55%
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Draw the structure of $Xe{F}_{2}$ molecule.

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

$Xe{F}_{2}$ :

Shape : Linear

Angle : $F-Xe-F>90°$

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Draw the structures of the following molecules :

(i) $Xe{F}_{4}$                                             (ii) $Br{F}_{3}$

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(i) $Xe{F}_{4}$

Shape : Square planar

(ii) $Br{F}_{3}$

Shape : T-shape

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Draw the structural formulae of molecules of following compounds.

(i) $Br{F}_{3}$ and                                  (ii) $Xe{F}_{4}$

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(i) $Br{F}_{3}$

Shape : T-shape

(ii) $Xe{F}_{4}$

Shape : Square planar

Difficulty Level:

• 82%
• 0%
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Explain the following :

(a) Xenon does not form such fluorides as $Xe{F}_{3}$ and $Xe{F}_{5}$.

(b) Out of noble gases, only Xenon is known to form real chemical compounds.

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(a) By impairing of one paired orbital, two singly occupied orbitals come into existence. Thus, either two or four or six singly occupied orbitals can be formed instead of one, three or five singly occupied orbitals. Hence $XeF$ are not formed.

(b) $Xe$ atom has a large size and lower ionisation potential and hence the force of nucleus over the electrons is weak and hence very small energy can excite the electrons and hence it is easier for Xenon to form compounds than other noble gases.

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How is $Xe{O}_{3}$ obtained? Write the related chemical equations. Draw the structure of $Xe{O}_{3}.$

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

Hydrolysis of $Xe{F}_{4}$ and $Xe{F}_{6}$ with water gives $Xe{O}_{3}$

$6Xe{F}_{4}+12{H}_{2}O\to 4Xe+2Xe{O}_{3}+24HF+3{O}_{2}$

$Xe{F}_{6}+3{H}_{2}O\to Xe{O}_{3}+6HF$

Shape : $Xe{O}_{3}$ is pyramidal structure

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Complete the following equations :

(i) C + conc.${H}_{2}S{O}_{4}\to$

(ii) $Xe{F}_{2}+{H}_{2}O\to$

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(i) $C+2{H}_{2}S{O}_{4}\left(conc.\right)\to C{O}_{2}+2S{O}_{2}+2{H}_{2}O$

(ii)

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• 91%
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How would you account for the following:

(i) $NC{l}_{3}$ is an endothermic compound while $N{F}_{3}$ is an exothermic one.

(ii) $Xe{F}_{2}$ is a linear molecule without a bend.

(iii) The electron gain enthalpy with negative sign for fluorine is less than that for chlorine, still fluorine is a stronger oxidising agent than chlorine.

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(i) F is more electronegative than Cl.

The difference in the electronegativity between N and F is much more than the difference between     electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.

(ii) In $Xe{F}_{2}$ there are 2 bond pairs and 3 lone pairs and thus show $s{p}^{3}d$ hybridization. It has linear       geometry.

(iii) Because of small size of flourine atom and strong electron-electron repulsions in its compact 2p        orbitals.

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(a) Account for the following :

(i) Helium is used in diving apparatus.

(ii) Fluorine does not exhibit positive oxidation state.

(iii) Oxygen shows catenation behaviour less than sulphur.

(b) Draw the structures of the following molecules :

(i) $Xe{F}_{2}$                                                (ii) ${H}_{2}{S}_{2}{O}_{8}$

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(a) (i) Helium is used in diving apparatus because of its very low solubility in blood and therefore an oxygen-helium mixture is used for artifical respiration.

(ii) Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.

(iii) The greater catenation tendency of sulphur is due to two reasons :

(a) The lone pair of electrons feels more repulsion in O-O bond than S-S bond due to its small size and thus S-S forms strong bond.

(b) As the size of atom increases down the group from O-PO, the strength of bond increases and therefore catenation tendency also increases.

(b) (i) $Xe{F}_{2}$ :

Shape : Linear

Angle : $F-Xe-F>90°$

(ii) ${H}_{2}{S}_{2}{O}_{8}$ :

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(a) Write the balanced chemical equations for obtaining $Xe{O}_{3}$ and $XeO{F}_{4}$ from $Xe{F}_{6}.$

(b) Account for the following :

(i) ${H}_{2}S$ is less acidic than ${H}_{2}Te$.

(ii)${H}_{3}P{O}_{2}$ has reducing nature.

(iii) $S{O}_{2}$ is an air pollutant.

Concept Videos :-

#8 | 18th Group

Concept Questions :-

group 18,perparation and properties

(a) $Xe{F}_{6}+3{H}_{2}O\to Xe{O}_{3}+6HF$

$Xe{F}_{6}+{H}_{2}O\to XeO{F}_{4}+2HF$

(b) (i) In the group with increase in size of the element and increased bond distance, the bond dissociation energy decreases, therefore H-S bond dissociation energy is higher than H-Te and hence H-S bond breaks less easily than H-Te bond and H2S is a weaker acid than ${H}_{2}Te$.

(ii) Because it contains two P-H bonds and thus reduces $AgN{O}_{3}$ to metallic silver

$4AgN{O}_{3}+{H}_{3}P{O}_{2}+2{H}_{2}O\to 4Ag↓+{H}_{3}P{O}_{4}+4HN{O}_{3}$

(iii) $S{O}_{2}$ is a pungent and irritating gas. It acts as an air pollutant due to the following reasons :

1) It causes throat and eye irritation as it is absorbed readily by respiratory tract.

2) It combines with moisture forming sulphurous acid. It is then converted into ${H}_{2}S{O}_{4}$. Both these acids cause acid rain and destroy the marble, corrode metals, deteriorate fabrics, paper, leather etc.

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