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1

AIPMT 2015 Cancelled Paper

MCQ (Single Correct Answer)
Nitrogen dioxide and sulphur dioxide have some properties in common . Which property is shown by one of these compounds , but not by the other?
A
Is soluble in water.
B
Is used as a food preservative.
C
Forms 'acid-rain'.
D
Is a reducing agent.

Explanation

SO2 is widely used in food and drinks industries for its property as a preservative and antioxidant while NO2 is not used as food preservative.
2

AIPMT 2015

MCQ (Single Correct Answer)
The formation of the oxide ion, O2$$-$$(g) from oxygen atom requires first an exthermic and then an endothermic step as shown below:
O(g) + e$$-$$ $$ \to $$ O$$-$$(g);  $$\Delta $$fHo = $$-$$141 kJ mol$$-$$1

O$$-$$(g) + e$$-$$ $$ \to $$ O2$$-$$(g);  $$\Delta $$fHo = +780 kJ mol$$-$$1

Thus, process of formation of O2$$-$$ is isoelectronic with neon. It is due to the fact that,
A
O$$-$$ ion has comparatively smaller size than oxygen atom
B
oxygen is more electronegative
C
addition of electron in oxygen results in larger size of the ion
D
electron repulsion outweighs the stability gained by achieving noble gas configuration.

Explanation

The process of formation of O2– in gas phase is unfavourable even though O2– is isoelectronic with neon because electron repulsion outweigh the stability gained by achieving noble gas configuration.
3

AIPMT 2015

MCQ (Single Correct Answer)
Strong reducing behavior of H3PO2 is due to
A
high electron gain enthalpy of phosphorus
B
high oxidation state of phosphorus
C
presence of two $$-$$OH groups and one P$$-$$H bond
D
presence of one $$-$$OH group and two P$$-$$H bonds.

Explanation

All oxyacids of phosphorus which have P—H bonds act as strong reducing agents. H3PO2 has two P—H bonds hence, it acts as a strong reducing agent.
4

AIPMT 2015

MCQ (Single Correct Answer)
The variation of the boiling points of the hydrogen halides is in the order HF > HI > HBr > HCl.

What explains the higher boiling point of hydrogen fluoride?
A
There is strong hydrogen bonding between HF molecules.
B
The bond energy of HF molecules is greater than in other hydrogen halides.
C
The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule.
D
The electronegativity of fluorine is much higher than for other elements in the group.

Explanation

HF forms strong intermolecular H-bonding due to high electronegativity of F. Hence, the boiling point of HF is exceptionally high. Boiling points of other hydrogen halides gradually increases from HCl to HI due to increase in the size of halogen atoms form Cl to I, which further increases the magnitude of van der Waal's forces.

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