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Chemistry NEET Questions — Free MCQ Practice

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Which of the following is most basic

A.

F^-

B.

Cl^-

C.

Br^-

D.

$I^{-}$

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Explanation

As we move from F^- to I^-, the size of the ion increases and hence electron density decreases from F^- to I^-. As a result,the tendency to attract proton decreases.Thus basic character decreases in the order F^- >Cl^->Br^- > I^-.

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Of the hybrides, $H_{2} O, H_{2} S, H_{2} S e a n d H_{2} T e$ , the one which has the smallest pK_a value is

A.

H₂O

B.

H₂S

C.

H₂Se

D.

H₂Te

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Explanation

Strongest acid possess smallest pK_a value.

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The shape of $C l O_{3}^{-}$ is

A.

Tetrahedral

B.

Triangular planer

C.

Triangular pyramid

D.

Planer

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An element (X) forms compounds of the formula $X C l_{3}, X_{2} O_{5} a n d C a_{3} X_{2},$ but does not form X $C l_{5}$ . Which of the following is the element

A.

B

B.

Al

C.

N

D.

P

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Explanation

N forms NCl₃, N₂O₅ and Ca₃N₂ . Nitrogen, due to no empty d- orbitals, cannot extend its covalency by more than three and hence do not form NCl₅.Due to its small size and high electronegativity it can accept 3 electrons to form $N^{3 -} i o n (C a_{3} N_{2})$

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Oxygen does not form OF₆ because

A.

It has a small size

B.

There are no vacant d- orbitals available

C.

It has high ionisation energy

D.

It has large size

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Explanation

Because of absence of d-orbitals, oxygen cannot increase its co-ordination number. Oxygen exhibits positive oxidation states of +1 in O₂F₂ and +2 in OF_2.

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The molecule having the smallest bond angle is

A.

H₂O

B.

H₂S

C.

H₂Te

D.

H₂Se

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Explanation

Bond angle decreases in the H₂O > H₂S > H₂Se > H₂Te.

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A yellow metallic powder is burnt in a stream of fluorine to obtain a colourless gas X which is thermally stable and chemically inert. Its molecule has octahedral geometry. Another colourless gas Y with same constituent atoms as that of X is obtained when sulphur dichloride is heated with sodium fluoride. Its molecule has triangular pyramidal structure X and Y are

A.

SF₄ and S₂F₂

B.

SF₆ and SF₄

C.

NaF and NaCl

D.

SF₄ and SF₆

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Explanation

$S + 3 F_{2} \to S F_{6} (X) c o l o u r l e s s g a s S C l_{2} + 4 N a F \to S F_{4} (Y) + S_{2} C l_{2} + 4 N a C l$

SF₄ has trigonal pyramidal shape due to sp³d hybridisation. SF₆ is octahedral due to sp³d² hybridisation. Sulphur has S₈ molecular geometry

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In nitrogen family the H-M-H bond angle in the hybrides MH₃ gradually becomes closer to 90^o on going from N to Sb. This shows that gradually

A.

The basic strength of hybrides increases

B.

Almost pure p-orbitals are used for M-H bonding

C.

The bond energies of M-H bond increases

D.

The bond pairs of electrons becomes nearer to the central atom

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Explanation

The hybrides have a pyramidal or tetrahedral shape with a lone pair of electrons in on of the orbitals.The H-M-H bond angle is less than the original 109^o28' tetrahedral bond angle(H-N-H in NH₃ is 106^o45') because of greater repulsion between lone pair and a bond pair than between two obnd pairs of electrons.Because electronegativity of M decreases from N to Bi, the bond pair lie farther away from the central atom, and the lone pair causes greater distortion of bond angle.

Thus H-P-H bond in PH₃ is 94^o , while in AsH₃ and SbH₃ it is about 91.8^o and 91.3^o respectively(closer to 90^o). This suggests that orbitals used for bonding are closer to pure p-orbitals.

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The volume of 2M NaOH required to convert 12 grams of NaH₂PO₄ [M=120gm mol^-1] into Na₃PO₄ is

A.

20 ml

B.

80 ml

C.

100 ml

D.

200 ml

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Explanation

The reaction is

$N a H_{2} P O_{4} + 2 N a O H \to N a_{3} P O_{4} + 2 H_{2} O n 120 g 2 \times 40 = 80 g (1 m o l e) (2 m o l e s) 120 g N a H_{2} P O_{4} r e q u i r e s 80 g N a O H 12 g N a H_{2} P O_{4} r e q u i r e s 8 g N a O H 80 g N a O H i s p r e s e n t i n 1 l i t r e i . e . 1000 m l o f 2 M N a O H 8 g N a O H i s p r e s e n t i n \frac{1000}{80} \times 8 = 100 m l o f 2 M N a O H$

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Assertion : In trimethyalamine (N(CH₃)₃), the nitrogen has a pyramidal geometry, wheres in trisilylamine (N(SiH₃)₃), it has a planer geometry.

Reason : In trimethylamine, nitrogen shows sp³ hybridization while in trisilyamine it shows sp² hybridization.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

In trimethylamine, nitrogen shows sp³ hybridization therefore trimethylamine has a pyramidal geometry while in trisilyamine it shows sp² hybridization resulting in a planer square.

Chemistry MCQs for NEET — Practice Questions with Answers

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