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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:

Trichloroacetaldehyde was subject to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :

A) Chloroform	B) 2, 2, 2-Trichloroethanol
C) Trichloromethanol	D) 2, 2, 2-Trichloropropanol

Answer & Explanation Answer: B) 2, 2, 2-Trichloroethanol

Explanation:

$\begin{matrix} C l \\ | \\ C l & - & C & - & C H O \\ | \\ C l \end{matrix}$ $\to_{C a n n i z z a r o' s R e a c t i o n}^{N a O H}$ $\begin{matrix} C l \\ | \\ C l & - & C & - & C O O N a \\ | \\ C l \end{matrix}$ $+$ $\begin{matrix} C l \\ | \\ C l & - & C & - & C H_{2} \\ | 2 & 1 \\ C l \end{matrix}$ $- O H$

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Q:

Silver Mirror test is given by which one of the following compounds?

1. Benzophenone

2. Acetaldehyde

3. Acetone

4. Formaldehyde

A) Only 1	B) Only 2
C) 1 and 3	D) 2 and 4

Answer & Explanation Answer: D) 2 and 4

Explanation:

Both Formaldehyde and Acetaldehyde will give this test

$H C H O \overset{[A g (N H_{3})_{2}^{+}]}{\to} \overset{A g}{S i l v e r m i r r o r} + O r g a n i c c o m p o u n d$

$C H_{3} - C H O \overset{[A g (N H_{3})_{2}^{+}]}{\to} \underset{S i l v e r m i r r o r}{A g} + O r g a n i c c o m p o u n d$

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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:

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :

A) 518 nm	B) 1035 nm
C) 325 nm	D) 743 nm

Answer & Explanation Answer: D) 743 nm

Explanation:

$\frac{1}{λ} = \frac{1}{λ 1} + \frac{1}{λ 2}$

$\frac{1}{355} = \frac{1}{680} + \frac{1}{λ 2}$

$\frac{1}{λ 2} = \frac{680 - 355}{680 \times 355}$

$\Rightarrow λ 2 = 743 n m$ .

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