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Q:

# Which of the following is called 'the King of Chemicals' ?

 A) Nitric acid B) Hydrochloric acid C) Silver nitrate D) Sulphuric acid

Explanation:

Sulphuric acid is called the King of Chemicals because it is used in the preparation of a very large number of other useful chemicals such as hydrochloric acid, nitric acid, dyes, drugs etc. In fact the annual production of shulpuric acid in a country is a measure of the economy of the country.

Subject: Chemistry
Q:

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

(1) $\inline \fn_jvn K_{2}O

(2) $\inline \fn_jvn Al_{2}O_{3}

(3) $\inline \fn_jvn MgO

(4) $\inline \fn_jvn Na_{2}O

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

Explanation:

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

$\inline \fn_jvn \therefore \xrightarrow[increasing\: basic\: strength]{Al_{2}O_{3}

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

$\inline \fn_jvn \therefore \xrightarrow[Increasing\: basic\: strength]{Na_{2}O

$\inline \fn_jvn \therefore Correct\: order\: is\: Al_{2}O_{3}

Filed Under: Chemistry
Exam Prep: AIEEE

6 229
Q:

Boron cannot form which one of the following anions?

(1) $\inline \fn_jvn BO_{2}^{-}$

(2) $\inline \fn_jvn BF_{6}^{3-}$

(3) $\inline \fn_jvn BH_{4}^{-}$

(4) $\inline \fn_jvn B(OH)_{4}^{-}$

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

Explanation:

Due to absence of low lying vacant d orbital in B, $\inline&space;\fn_jvn&space;sp^{3}d^{2}$ hybridization is not possible hence $\inline&space;\fn_jvn&space;BF_{6}^{3-}$will not formed.

Filed Under: Chemistry
Exam Prep: AIEEE

2 321
Q:

Among the following the maximum covalent character is shown by the compound
(1)$\inline \fn_jvn MgCl_{2}$
(2)$\inline \fn_jvn FeCl_{2}$
(3)$\inline \fn_jvn SnCl_{2}$
(4)$\inline \fn_jvn AlCl_{3}$

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

Explanation:

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

Filed Under: Chemistry
Exam Prep: AIEEE

1 301
Q:

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

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

Explanation:

$\inline&space;\fn_jvn&space;H^{+}-e^{-}\rightarrow&space;\frac{1}{2}H_{2}$

Apply Nernst equation

$\inline&space;\fn_jvn&space;E=o-\frac{0.059}{1}\log\frac{PH_{2}^{\frac{1}2{}}}{[H^{+}]}$

$\inline&space;\fn_jvn&space;E=-\frac{0.059}{1}\log&space;\frac{2^{\frac{1}{2}}}{1}$

Therefore E is Negative

Filed Under: Chemistry
Exam Prep: AIEEE

2 317
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

$\begin{matrix}&space;&&space;&&space;Cl&space;&&space;&&space;\\&space;&&space;&&space;|&&space;&&space;\\&space;Cl&&space;-&space;&C&space;&&space;-&space;&CHO&space;\\&space;&&space;&&space;|&&space;&&space;\\&space;&&space;&&space;Cl&&space;&&space;\end{matrix}\xrightarrow[Cannizaros&space;\:&space;reaction]{NaOH}\begin{matrix}&space;&&space;&&space;Cl&space;&&space;&&space;\\&space;&&space;&&space;|&&space;&&space;\\&space;Cl&&space;-&space;&C&space;&&space;-&space;&COONa&space;\\&space;&&space;&&space;|&&space;&&space;\\&space;&&space;&&space;Cl&&space;&&space;\end{matrix}+\begin{matrix}&space;&&space;&&space;Cl&space;&&space;&&space;\\&space;&&space;&&space;|&&space;&&space;\\&space;Cl&&space;-&space;&C&space;&&space;-&space;&CH_{2}&-&OH&space;\\&space;&&space;&&space;|2&&space;&1&space;\\&space;&&space;&&space;Cl&&space;&&space;\end{matrix}$