Chemistry Test - 33

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Chemistry Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following defines enthalpy of solution?

A
$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q)$

B
$N H_{4} C l (s) \to N H_{4}^{+} (a q) + C l^{-} (a q)$

C
$C u S O_{4} \cdot 5 H_{2} O (s) + H_{2} O \to C u S O_{4} (a q)$

D
$C u S O_{4} (s) + 5 H_{2} O \to C u S O_{4} \cdot 5 H_{2} O (s)$

Solution

\begin{aligned}
&H_{2} O(l) \rightarrow H^{+}(a q)+O H-(a q) \text { Enthalpy of dissociation }\\
&N H_{4} C l(s) \rightarrow N H_{4}^{+}(a q)+C l^{-}(a q) \text { Enthalpy of dissociation }\\
&\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CuSO}_{4}(a q) \text { Enthalpy of solution }\\
&C u S O_{4}(s)+5 H_{2} O \rightarrow C u S O_{4} \cdot 5 H_{2} O(s) \text { Enthalpy of hydration }
\end{aligned}
Question 2
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Urea and hydrazine react to form ammonia gas along with compound X which reacts with aldehydes and ketones to form specific crystalline derivatives. X is

A
phenyl hydrazine

B
semicarbazide

C
biuret

D
acetylurea

Solution

$N H_{2} C O N H_{2} + N H_{2} - N H_{2} \to N H_{2} N H C O N H_{2} + N H_{3}$
Question 3
1 / -0

A current of 2.0A when passed for 5 hours through a molten metal salt deposits 22.2g of metal of atomic weight 177.The oxidation state of the metal in the metal salt is

A
+1

B
+2

C
+3

D
+4

Solution
Question 4
1 / -0

The z - component of angular momentum of an electron in an atomic orbital is governed by the

A
azimuthal quantum number

B
principal quantum number

C
magnetic quantum number

D
spin quantum number

Solution

The expression is $L_{2}=m\left(\frac{h}{2 \pi}\right)$
Question 5
1 / -0

A metal X on heating in nitrogen gas gives Y.Y on treatment with H₂O gives a colourless gas which when passed through CuSO₄ solution gives a blue colour. Y is

A
Mg(NO₃)₂

B
Mg₃N₂

C
NH₃

D
MgO

Solution

$\begin{matrix} 3 M x g + N_{2} ≜ M g_{3} N_{2}; C u S O_{4} + 4 N H_{3} \to [C u {(N H_{3})}_{4}] S O_{4} \end{matrix}$
Question 6
1 / -0

The solubility of the halogen 'X' in water is greatly increased by the addition of its salt. The halogen' X' is

A
fluorine

B
chlorine

C
bromine

D
iodine

Solution

Solubility of I₂ is greatly increased by adding its salt KI in water solution.
Question 7
1 / -0

Sometimes yellow turbidity appears while passing H₂S gas even in the absence of II group radicals. This is because of

A
sulphur is present in the mixture as impurity

B
IV group radicals are precipitated as sulphides

C
the oxidation of H2Sgas by some acid radicals

D
III group radicals are precipitated as hydroxides

Solution

Turbidity is due to colloidal formed, as a result of the oxidation of $H_{2} S$ by some oxidising agent (radicals) present in the mixture. $H_{2} S+[O] \rightarrow H_{2} O+S$
Question 8
1 / -0

Diborane has

A
2 two - centre - two - electron bonds and 4 three - centre - three - electron bonds.

B
4 two - centre - two - electron bonds and 2 three - centre - three - electron bonds

C
3 two - centre - two - electron bonds and 3 three - centre - three - electron bonds.

D
all six identical bonds.

Solution

Structure of $B_{2} H_{6}$ is as follows:

Thus the diborane

molecule has four two-centre-two- electron bonds (2c-2e bonds) also called usual
bonds and two three-centre-two-electron bonds (3c?2e) also called banana bonds.
Hydrogen attached to usual and banana bonds are called $\mathrm{H}_{\mathrm{t}}$ (terminal $\mathrm{H}$ ) and
$\mathrm{H}_{\mathrm{b}}($ bridgedH) respectively.
Question 9
1 / -0

Which of the following has a shape different from others?

A
[Zn(NH₃)₄]⁺²

B
Ni(CO)₄

C
[Cd(CN)₄]⁻²

D
[Cu(NH₃)₄]⁺²

Solution

$\left[Z n\left(N H_{3}\right)_{4}\right]^{2+}, N i(C O)_{4} \&\left[C d(C N)_{4}\right]^{2-}$ all form tetrahedral structures where the central atom uses sp3 hybridisation. Whereas $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ uses dsp 2
hybridisation and forms square planar structure. Although Cu should use both
tetrahedral (sp3) \& square planar (dsp2) hybridisations but the later one is confirmed
through X-ray analysis.
Question 10
1 / -0

In which of the following cases, the stability of two oxidation states is correctly represented

A
Ti³⁺ > Ti⁴⁺

B
Mn²⁺ > Mn³⁺

C
Fe²⁺ > Fe³⁺

D
Cu⁺ > Cu²⁺

Solution

Mn²⁺(3d⁵) is more stable then Mn³⁺(3d⁴). due to half-filled orbitals

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Chemistry Test - 33

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Chemistry Test - 33

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Question 1

H2O(l)→H+(aq)+OH−(aq)

NH4Cl(s)→NH4+(aq)+Cl−(aq)

CuSO4⋅5H2O(s)+H2O→CuSO4(aq)

CuSO4(s)+5H2O→CuSO4⋅5H2O(s)

Solution

Question 2

phenyl hydrazine

semicarbazide

biuret

acetylurea

Solution

Question 3

+1

+2

+3

+4

Solution

Question 4

azimuthal quantum number

principal quantum number

magnetic quantum number

spin quantum number

Solution

Question 5

Mg(NO3)2

Mg3N2

NH3

MgO

Solution

Question 6

fluorine

chlorine

bromine

iodine

Solution

Question 7

sulphur is present in the mixture as impurity

IV group radicals are precipitated as sulphides

the oxidation of H2Sgas by some acid radicals

III group radicals are precipitated as hydroxides

Solution

Question 8

2 two - centre - two - electron bonds and 4 three - centre - three - electron bonds.

4 two - centre - two - electron bonds and 2 three - centre - three - electron bonds

3 two - centre - two - electron bonds and 3 three - centre - three - electron bonds.

all six identical bonds.

Solution

Question 9

[Zn(NH3)4]+2

Ni(CO)4

[Cd(CN)4]−2

[Cu(NH3)4]+2

Solution

Question 10

Ti3+ > Ti4+

Mn2+ > Mn3+

Fe2+ > Fe3+

Cu+ > Cu2+

Solution

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$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q)$

$N H_{4} C l (s) \to N H_{4}^{+} (a q) + C l^{-} (a q)$

$C u S O_{4} \cdot 5 H_{2} O (s) + H_{2} O \to C u S O_{4} (a q)$

$C u S O_{4} (s) + 5 H_{2} O \to C u S O_{4} \cdot 5 H_{2} O (s)$

Mg(NO₃)₂

Mg₃N₂

NH₃

[Zn(NH₃)₄]⁺²

Ni(CO)₄

[Cd(CN)₄]⁻²

[Cu(NH₃)₄]⁺²

Ti³⁺ > Ti⁴⁺

Mn²⁺ > Mn³⁺

Fe²⁺ > Fe³⁺

Cu⁺ > Cu²⁺