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Chemistry Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following ions has the lowest magnetic moment ?

A
Cu²⁺

B
Ni²⁺

C
Co³⁺

D
Fe²⁺

Solution

Greater the number of unpaired electrons, greater will be the magnetic moment. Let us consider the electronic configuration of the options:-
A) $C u^{2+}-1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{5} 3 d^{9}-1$ unpaired electron.
B) $N i^{2+}-1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6} 3 d^{8}-2$ unpaired electrons.
C) $C o^{3+}-1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6} 3 d^{6}-4$ unpaired electrons.
D) $F e^{2+}-1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6} 3 d^{6}-4$ unpaired electrons.

Hence option A is the right answer.
Question 2
1 / -0

The set(s) representing the incorrect order of ionic radius is/are:

A
Li⁺>Be²⁺>Na⁺>Mg²⁺

B
Na⁺>Mg²⁺>Be²⁺

C
Na⁺>Be²⁺

D
Na⁺>Mg²⁺>Li⁺>Be²⁺

Solution

The set(s) representing the incorrect order of ionic radius is, $L i^{+}>B e^{2+}>N a^{+}>M g^{2+} .$ The correct order is, $N a^{+}>M g^{2+}>L i^{+}>B e^{2+} .$ On moving down the group, the atomic and ionic radii increases.

Hence option A is the correct answer.
Question 3
1 / -0

What is the total number of valence electrons in the peroxydisulphate, $S_{2} O_{8}^{2-},$ ion?

A
58

B
60

C
62

D
64

Solution

The total number of valence electrons in the peroxydisulphate ion $S_{2} O_{8}^{2-},$ is 62.
$2(6)+8(6)+2=12+48+2=62$

Hence option C is the correct answer.
Question 4
1 / -0

Atomic radii of fluorine and neon in angstrom units are, respectively, given by :

A
0.72, 1.60

B
1.60, 1.60

C
0.72, 0.72

D
1.60, 0.72

Solution

Florine has 5 electrons in $2 p$ orbital which is less than what is required for ideal configuration
inert gas configuration. Hence nuclear charge is high leading smaller size than the next inert
gas neon.
Hence atomic radius of fluorine and neon are 0.75 angstrom and 1.6 angstrom.

Hence option A is the correct answer.
Question 5
1 / -0

Select the correct order of 2nd ionization energy of C, N, O, and F.

A
C > N > O > F

B
N > C > O > F

C
C < N < O < F

D
O > F > N > C

Solution

Consider the configurations of the given elements after the loss of an electron.

C+ - 1s²2s²2p¹

N+ - 1s²2s²2p²

O+ - 1s²2s²2p³

F+ - 1s²2s²2p⁴

We notice that O has a half filled orbital and hence has the highest I.E. among these. Also, carbon can lose an electron easily. Flourine will reach a half filled configuration upon the loss of an electron.

Hence, the order is O>F>N>C.
Hence, the correct option is (D)
Question 6
1 / -0

The incorrect hydrated radius order is:

A
$L i_{(a q)}^{+} < B e_{(a q)}^{2 +}$

B
$N a_{(a q)}^{+} < A l_{(a q)}^{3 +}$

C
$I^{-} (a q) > C l^{-} (a q)$

D
$B a_{(a q)}^{2 +} < C a_{(a q)}^{2 +}$

Solution

The incorrect hydrated radius order is I(aq¯¯¯¯¯)>Cl(aq).
The correct order of hydrated radius is $I_{a q}$ Higher is the charge and lower is the size, higher is the extent of hydration and higher is the hydrated radius.

Hence option C is the correct answer.
Question 7
1 / -0

Select equations having exothermic step.
(I) $S_{(g)}^{-} \rightarrow S_{(g)}^{2-}$
$( II ) N a_{(g)}^{+}+C l_{(g)}-\rightarrow N a C l_{(s)}$
$($ III $) N _{(g)} \rightarrow N _{(g)}$
(IV) $A l_{(g)}^{2+} \rightarrow A l_{(g)}^{3+}$
Choose the correct code(s).

A
II

B
I,II

C
III Aand IV

D
II and III

Solution

a. $\mathrm{S}_{(\mathrm{g})}^{-} \rightarrow \mathrm{S}_{(\mathrm{g})}^{2-}$ Electron affinity of an anion (endothermic)

b. $\mathrm{Na}_{(\mathrm{g})}^{+}+\mathrm{Cl}_{(\mathrm{g})}^{-} \rightarrow \mathrm{NaCl}_{(\mathrm{s})}$ Lattice energy (exothermic)

c. $\mathrm{N}_{(\mathrm{g})} \quad \rightarrow \quad \mathrm{N}_{(\mathrm{g})}^{-}$

More stable $\quad$ Less stable

(Half filled p-subshell)

d. $\mathrm{Al}_{(\mathrm{g})}^{2+} \rightarrow \mathrm{Al}^{3+}(\mathrm{g})$ (endothermic as ionization is endothermic)

Hence option A is the correct answer.
Question 8
1 / -0

Which terms are exothermic for the formation of NaF(s)?
$I \cdot N a_{(g)} \rightarrow N a_{(g)}^{+}+e^{-}$
II. $F_{(g)}+e^{-} \rightarrow F_{(g)}$
$\operatorname{III.} N a_{(g)}^{+}+F_{(\bar{g})} \rightarrow N a F_{(s)}$

A
I only

B
II only

C
I and III only

D
II and III only

Solution

The first term indicates the ionization energy which is endothermic and has positive value.

The second term indicates electron affinity which is exothermic and has negative value.

The third term indicates lattice enthalpy which is exothermic and has negative value.

Hence option D is the correct answer.
Question 9
1 / -0

Which configuration represents atom having the highest second ionization energy ?

A
1s²,2s²2p⁴

B
1s²,2s²2p⁶

C
1s²,2s²2p⁶,3s¹

D
1s²,2s²2p⁶,3s²

Solution

Consider the configurations after the loss of one electron:-
A) $1 s^{2}, 2 s^{2} 2 p^{3}$
B) $1 s^{2}, 2 s^{2} 2 p^{5}$
C) $1 s^{2}, 2 s^{2} 2 p^{6}$
D) $1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{1}$
From the above we notice that option $C$ has attained noble gas configuration. Hence it will have the highest ionization energy compared to the rest. T

Hence option $C$ is the right answer.
Question 10
1 / -0

From the electronic configuration of the given element K, L, M and N, which one has the highest ionization potential:

A
K=[Ne]3s²,3p²

B
L=[Ne]3s²,3p³

C
M=[Ne]3s²,3p¹

D
N=[Ar]3d¹⁰,4s²,4p³

Solution

We know that half filled and fully filled configurations have more ionization energy. Hence from the options we notice that B and D have half filled orbitals. They therefore have greater Ionization energy compared to A and C.

Among B and D we notice that the half filled orbital in D is the 4^th orbit whereas for B it is in the 3^rd. Therefore due to shielding effect, the ionization energy of option B will be greater.

Hence option B is the correct answer.

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Chemistry Test - 10

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Question 1

Cu2+

Ni2+

Co3+

Fe2+

Solution

Question 2

Li+>Be2+>Na+>Mg2+

Na+>Mg2+>Be2+

Na+>Be2+

Na+>Mg2+>Li+>Be2+

Solution

Question 3

58

60

62

64

Solution

Question 4

0.72, 1.60

1.60, 1.60

0.72, 0.72

1.60, 0.72

Solution

Question 5

C > N > O > F

N > C > O > F

C < N < O < F

O > F > N > C

Solution

Question 6

Li(aq)+<Be(aq)2+

Na(aq)+<Al(aq)3+

I−(aq)>Cl−(aq)

Ba(aq)2+<Ca(aq)2+

Solution

Question 7

II

I,II

III Aand IV

II and III

Solution

Question 8

I only

II only

I and III only

II and III only

Solution

Question 9

1s2,2s22p4

1s2,2s22p6

1s2,2s22p6,3s1

1s2,2s22p6,3s2

Solution

Question 10

K=[Ne]3s2,3p2

L=[Ne]3s2,3p3

M=[Ne]3s2,3p1

N=[Ar]3d10,4s2,4p3

Solution

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Cu²⁺

Ni²⁺

Co³⁺

Fe²⁺

Li⁺>Be²⁺>Na⁺>Mg²⁺

Na⁺>Mg²⁺>Be²⁺

Na⁺>Be²⁺

Na⁺>Mg²⁺>Li⁺>Be²⁺

$L i_{(a q)}^{+} < B e_{(a q)}^{2 +}$

$N a_{(a q)}^{+} < A l_{(a q)}^{3 +}$

$I^{-} (a q) > C l^{-} (a q)$

$B a_{(a q)}^{2 +} < C a_{(a q)}^{2 +}$

1s²,2s²2p⁴

1s²,2s²2p⁶

1s²,2s²2p⁶,3s¹

1s²,2s²2p⁶,3s²

K=[Ne]3s²,3p²

L=[Ne]3s²,3p³

M=[Ne]3s²,3p¹

N=[Ar]3d¹⁰,4s²,4p³