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Q:

Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?

(1) $K_{2} O < N a_{2} O < A l_{2} O_{3} < M g O$

(2) $A l_{2} O_{3} < M g O < N a_{2} O < K_{2} O$

(3) $M g O < K_{2} O < A l_{2} O_{3} < N a_{2} O$

(4) $N a_{2} O < K_{2} O < M g O < A l_{2} O_{3}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: B) Option 2

Explanation:

While moving from left to right in periodic table basic character of oxide of elements will decrease.

$∴ \to_{I n c r e a \sin g b a s i c s t r e n g t h}^{A l_{2} O_{3} < M g O < N a_{2} O}$

and while descending in the group basic character of corresponding oxides increases.

$∴ \to_{I n c r e a \sin g b a s i c s t r e n g t h}^{N a_{2} O < K_{2} O}$

$∴ C o r r e c t O r d e r i s A l_{2} O_{3} < M g O < N a_{2} O < K_{2} O$

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Q:

Boron cannot form which one of the following anions?

(1) $B O_{2}^{-}$

(2) $B F_{6}^{3 -}$

(3) $B H_{4}^{-}$

(4) $B (O H)_{4}^{-}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: B) Option 2

Explanation:

Due to absence of low lying vacant d orbital in B, $s p^{3} d^{2}$ hybridization is not possible hence $B F_{6}^{3 -}$ will not formed.

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Q:

Among the following the maximum covalent character is shown by the compound

(1) $M g C l_{2}$

(2) $F e C l_{2}$

(3) $S n C l_{2}$

(4) $A l C l_{3}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: D) Option 4

Explanation:

According to Fajans rule, cation with greater charge and smaller size favours covalency.

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Q:

The reduction potential of hydrogen half-cell will be negative if :
(1) $P (H_{2})$ = 2 atm and [H+] = 2.0 M
(2) $P (H_{2})$ = 1 atm and [H+] = 2.0 M
(3) $P (H_{2})$ = 1 atm and [H+] = 1.0 M
(4) $P (H_{2})$ = 2 atm and [H+] = 1.0 M

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: D) Option 4

Explanation:

$H^{+} - e^{-} \to \frac{1}{2} H_{2}$

Apply Nernst equation

$E = 0 - \frac{0.059}{1} \log (\frac{P H_{2}^{\frac{1}{2}}}{[H^{+}]})$

$E = - \frac{0.059}{1} \log (\frac{2^{\frac{1}{2}}}{1})$

Therefore E is Negative

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Q:

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 $d m^{3}$ at 27°C is to a volume of 100 $d m^{3}$

A) 42.3 J/ mole / K	B) 38.3 J/ mole / K
C) 35.8 J/ mole / K	D) 32.3 J/ mole / K

Answer & Explanation Answer: B) 38.3 J/ mole / K

Explanation:

$∆ s = n R \ln \frac{v 2}{v 1}$

$= 2.303 n R \log \frac{v 2}{v 1}$ = $2.303 \times 2 \times 8.314 \times \log (\frac{100}{10})$ = $38.3 J$ $m o l^{- 1}$ $k^{- 1}$

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Q:

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is

1) $A_{2} B_{5}$

2) $A_{3} B_{2}$

3) $A_{2} B_{3}$

4) $A_{3} B_{4}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: A) Option 1

Explanation:

$Z_{A} = \frac{8}{8}$ ; $Z_{B} = \frac{5}{2}$

So formula of compound is $A B_{\frac{5}{2}}$

i.e., $A_{2} B_{5}$ .

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Q:

The degree of dissociation ( $α$ ) of a weak electrolyte, $A_{x} B_{y}$ is related to van't Hoff factor (i) by the expression:

$1 . α = \frac{x + y + 1}{i - 1}$

$2 . α = \frac{x + y - 1}{i - 1}$

$3 . α = \frac{i - 1}{x + y + 1}$

$4 . α = \frac{i - 1}{x + y - 1}$

A) 1 is correct	B) 2 is correct
C) 3 is correct	D) 4 is correct

Answer & Explanation Answer: D) 4 is correct

Explanation:

Van't Hoff factor (i) = $\frac{O b s e r v e d C o l l i g a t i v e P r o p e r t y}{N o r m a l C o l l i g a t i v e P r o p e r t y}$

$\begin{matrix} A_{x} B_{y} \\ 1 - α \end{matrix} \to \begin{matrix} x A^{+ y} \\ x α \end{matrix} + \begin{matrix} y B^{- x} \\ y α \end{matrix}$

Total Moles = $1 - α + x α + y α$ = $1 - α (x + y - 1)$

$i = \frac{1 + α (x + y - 1)}{1}$

$\Rightarrow α = \frac{i - 1}{x + y - 1}$

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Q:

The structure of $I F^{7}$ is

A) Pentagonal bipyramid	B) Square pyramid
C) Trigonal bipyramid	D) Octahedral

Answer & Explanation Answer: A) Pentagonal bipyramid

Explanation:

Hybridisation of iodine is $s p^{3} d^{3}$

So, the structure is pentagonal bipyramid.

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