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Chemistry MCQs for NEET — Practice Questions with Answers

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In the equation

$H_{2} S + 2 H N O_{3} \to 2 H_{2} O + 2 N O_{2} + S$

The equivalent weight of hydrogen sulphide is [BVP 2003]

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Explanation

$H_{2} S \to \overset{0}{S} + 2 e$

Equivalent wt. = $\frac{Mol . wt .}{2}$ = $\frac{34}{2}$ = 17.

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If 1.2 g of metal displace 1.12 litre hydrogen at normal temperature and pressure, equivalent weight of metal would be [DPMT 2001]

1.2 $\div$ 11.2

1.2 $\times$ 11.2

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Explanation

$1 .12 l t r H_{2} = 1 .2 g;$

$∴ 22 .4 l t r H_{2} = 24 g$

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The equivalent weight of phosphoric acid (H₃PO₄) in the reaction $N a O H + H_{3} P O_{4} \to N a H_{2} P O_{4} + H_{2} O$ is [AIIMS 1999]

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Explanation

Molecular weight of H₃PO₄ is 98 and change in

its valency = 1 equivalent wt. of H₃PO₄

$= \frac{Molecular weight}{Change in valency} = \frac{98}{1} = 98$ .

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What is the equivalent mass of $I O_{4}^{-}$ when it is converted into I₂ in acid medium [Kerala PMT 2004]

M/6

M/7

M/5

M/4

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Explanation

Equivalent mass

$= \frac{Molecular weight}{Change in oxidation number per mole}$

Suppose molecular weight is M

Oxidation number of I₂ in $I O_{4}^{-}$ in Acidic medium i.e., $I \times (- 8) + 1 e^{-} = + 7$

So eq. wt. = M/7.

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In alkaline medium ClO₂ oxidize H₂O₂ in O₂ and reduced itself in Cl^– then how many mole of H₂O₂ will oxidize by one mole of ClO₂

1.0

1.5

2.5

3.5

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Explanation

$C l O_{2} \to C l^{-}$

$C l O_{2} + 2 H_{2} O + 5 e \to C l^{-} + 4 O H^{-}$

$H_{2} O_{2} \to O_{2}$

$H_{2} O_{2} + 2 O H^{-} \to O_{2} + 2 H_{2} O + 2 e$

$C l O_{2} + 2 H_{2} O + 5 e \to C l^{-} + 4 O H^{-}] \times 2$

$H_{2} O_{2} + 2 O H^{-} \to O_{2} + 2 H_{2} O + 2 e] \times 2$

$2 C l O_{2} + 5 H_{2} O_{2} + 2 O H^{-} \to 2 C l^{-} + 5 O_{2} + 5 H_{2} O$

$2 C l O_{2} \equiv 5 H_{2} O_{2}$

∴ $C l O_{2} = 2 .5 H_{2} O_{2}$

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Oxidation number of oxygen in potassium super oxide (KO₂) is [UPSEAT 1999, 2002]

– 2

– 1

– 1/2

– 1/4

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Explanation

$\overset{*}{K O_{2}}$ , $+ 1 + 2 x = 0, x = - \frac{1}{2}$ .

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The compound ${YBa}_{2} {Cu}_{3} O_{7}$ which shows superconductivity has copper in oxidation state ........ Assume that the rare earth element Yttrium is in its usual +3 oxidation state [IIT 1994]

3/7

7/3

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Explanation

$B a_{2} \overset{*}{C u_{3}} O_{7}$

$3 + 2 \times 2 + 3 x - (2 \times 7) = 0$

$3 + 4 + 3 x - 14 = 0$

$3 x = 7$

$x = \frac{7}{3}$ .

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In a balanced equation $H_{2} S O_{4} + x H I \to H_{2} S + y I_{2} + z H_{2} O$ , the values of x, y, z are [EAMCET 2003]

x = 3, y = 5, z = 2

x = 4, y = 8, z = 5

x = 8, y = 4, z = 4

x = 5, y = 3, z = 4

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Explanation

The values of x, y, z are 8, 4, 4 respectively hence the reaction is

$H_{2} S O_{4} + 8 H I \to H_{2} S + 4 I_{2} + 4 H_{2} O$

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When $M n O_{2}$ is fused with KOH, a coloured compound is formed, the product and its colour is [IIT Screening 2003]

$K_{2} M n O_{4}$ , purple green

$K M n O_{4}$ , purple

$M n_{2} O_{3}$ , brown

$M n_{3} O_{4}$ black

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Explanation

$2 M n O_{2} + 4 K O H + O_{2} \overset{Δ}{\to} \underset{purple green}{2 K_{2} M n O_{4}} + 2 H_{2} O$ .

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In the reaction, $P + NaOH \to {PH}_{3} + {NaH}_{2} {PO}_{2}$

P is oxidized only

P is reduced only

P is oxidized as well as reduced

Na is reduced

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Explanation

P goes from an oxidation number

0 to + 1-- oxid

0 to -3 -- red

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