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In the following reaction:

4P + 3KOH + 3H₂O $\to$ 3KH₂PO₂ + PH₃:

P oxidised only

P is reduced only

P is oxidised as well as reduced

none of the above

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Explanation

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In which SO₂acts as oxidant, while reacting with:

acidified KMnO₄

acidified K₂Cr₂O₇

H₂S

acidified C₂H₅OH

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Explanation

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In the reaction;

3Br₂ + 6C $O_{3}^{2 -}$ + 3H₂O $\to$ 5Br^- + Br $O_{3}^{-}$ + 6HC $O_{3}^{-}$

which statement is correct?

Br₂ is oxidised

Br₂ is reduced

Br₂ is neither oxidised nor reduced

Br₂ is oxidised and reduced as well

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Explanation

(d) B $r_{2}^{0}$ $\to$ 2Br⁵⁺ + 10e; B $r_{2}^{0}$ + 2e $\to$ 2Br^-

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When NaCl is dissolved in water, the sodium ion becomes:

oxidised

reduced

hydrolysed

hydrated

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Explanation

(d) Ions are hydrated on dissolution of salt in water.

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Hypochlorite disproportionate to give

Cl^- and Cl $O_{4}^{-}$

Cl $O_{4}^{-}$ and Cl $O_{3}^{-}$

Cl $O_{3}^{-}$ and Cl^-

Cl $O_{2}^{-}$ and Cl^-

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Which has highest normality?

1MH₃PO₂

1MH₃PO₃

1MH₃PO₄

All of these

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Explanation

H₃PO₂ is monobasic, H₃PO₃ is diabasic and H₃PO₄ is tribasic.

1MH₃PO₂ =1NH₃PO₃ , 1M H₃PO₃ =2N H₃PO₃ and 1M H₃PO₄ =3N H₃PO₄

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Amongst the following identify the species with an atom in (+6) oxidation state.

Mn $O_{4}^{-}$

Cr $(CN)_{}^{3 -} 6$

Ni $F_{6}^{2 -}$

CrO₂Cl₂

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Explanation

Cr O₂ Cl₂

x + 2(-2) + 2(-1) = 0

x = +6

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Mg(OH)₂ + HCl $\to$ Mg(OH)Cl + H₂O₂, the equivalent weight of Mg(OH)₂ will be (M= Mol. wt. of Mg(OH)₂)

zero

M/2

M/4

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Explanation

Mg(OH)₂ + HCl $\to$ Mg(OH)Cl + H₂O₂,

In this reaction, only one -OH group of Mg(OH)₂ is replaced by -Cl. Hence, n-factor of Mg(OH)₂ will be 1.

Equivalent wt. of Mg(OH)₂ = M/l=M

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A metallic oxide contains 20% oxygen. The equivalent weight of metal is

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Explanation

In metallic oxide, 80% will be metal Suppose, long wt. of metal = 80g

wt. of oxygen = 20g

Equivalents of metal = Equivalents of oxygen

$\frac{80}{E_{M}} = \frac{20}{8}$

E_M = 32

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2 gm limestone (CaCO₃ + Impurities ) is reacted with 100ml N/2 HCl. The excess acid requires 60ml N/2 NaOH. The % purity of limestone is

25%

50%

75%

100%

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Explanation

Total milliequivalents of HCl = 100x1/2=50

Left milliequivalents of HCl = 60x1/2=30

Used milliequivalents of HCl = 20

Miliequivalents of pure CaCO₃ = Milli w/50x1000=20

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