Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Halogen with highest electron affinity is:

A
F

B
Cl

C
Br

D
I

Solution

Fluorine, though higher electronegativity than chlorine in the periodic table, has a very small atomic size.

This makes the fluoride anion so formed unstable due to a very high charge/mass ratio.

Also, fluorine has no d electrons which limits its atomic size.

As a result, fluorine has an electron affinity less than that of chlorine.

Option B is correct.
Question 2
1 / -0

Which is not correct in case of $$Be$$ and $$Al$$?

A
Both are rendered passive by conc. $$HNO_{3}$$

B
Carbides of both give methane on hydrolysis

C
Both give hydroxides which are basic

D
Both give covalent chlorides

Solution

Both form amphoteric hydroxide/oxide not either basic or acidic.

Because as Aluminium is in 3rd group it can either accept electron or donate hence they form oxides which are amphoteric.

The reason why beryllium forms is aluminium and beryllium has diagonal relationship.

Hence both will have common properties.

Option C is correct.
Question 3
1 / -0

The alkaline earth metal which shows properties similar to aluminium is:

A
$$Ca$$

B
$$Be$$

C
$$Sr$$

D
$$Ba$$

Solution

Beryllium and aluminum shows diagonal relationship with each other.

Hence, they exhibit similar properties.

Option B is correct.
Question 4
1 / -0

Which of the following is the correct order of chemical reactivity with water according to electrochemical series ?

A
$$K > Mg > Zn > Cu$$

B
$$K > Mg > Cu > Zn$$

C
$$Cu > Zn > K > Mg$$

D
$$Mg > K > Cu > Zn$$

Solution

The correct order for the chemical reactivity with water is $$K > Mg > Zn > Cu$$. This is because, among the given metals, potassium has the most negative standard reduction potential and copper has the highest standard reduction potential.

Metal
$$K$$ $$Mg$$

$$Zn$$ $$Cu$$

Standard reduction
potential (V)
$$-2.95$$ $$-2.363$$
$$-1.662$$ $$+0.340$$

Option A is correct.
Question 5
1 / -0

Among the following groupings, which represents the collection of isoelectronic species?

A
$$CO,NO^{+} ,CN^{-} ,C^{2-}_{2}$$

B
$$N_{2} ,C_{2}^{2-} ,CO, NO$$

C
$$NO,CN , N_{2} ,O_{2}^{-}$$

D
$$NO^{+} ,C^{2-}_{2} ,O_{2}^{-} ,CO$$

Solution

In option $$A$$, each species contains 14 electrons
$$CO =$$ 6(carbon) + 8(oxygen) $$= 14$$
$$NO^{+}=$$ 7(nitrogen) + 8(oxygen) - 1(the positive charge) $$= 14$$
$$CN^{-}=$$ 6(carbon) + 7(nitrogen) + 1(the negative charge) $$= 14$$
$$C_{2}^{-2}=$$ 6(carbon) + 6(carbon) + 2(-2 negative charge) $$= 14$$
So, $$A$$ is the correct answer.
Question 6
1 / -0

The electronic configurations of three elements X, Y and Z are given below. Which of the following is correct?
X: 2, 8
Y: 2, 8, 7
Z: 2, 8, 2

A
X is a metal

B
Y is a metal

C
Z is a non-metal

D
Y is a non-metal and Z is a metal

Solution

Generally, metals have a lesser number of electrons in their outermost shell which can be a maximum of three. So, from the given compounds Y cannot be a metal and X has a complete electronic configuration in the valence shell and is not a metal. When the number of valence electrons is less than or equal to 4, the valency is equal to valence electrons and when valence electrons > 4, then valency $$= (8-4)$$.
X- 2, 8; is a noble gas (Ne) with a valency of 0.
Y-2, 8, 7; is a non-metal or halogen (Cl) with a valency of $$8-7 = 1$$.
Z-2, 8, 2; is a metal (Mg) with a valency of 2.
Question 7
1 / -0

Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?

A
$$Cl < F < S < O$$

B
$$O < S < F < Cl$$

C
$$S < O < Cl < F$$

D
$$F < Cl < O < S$$
Question 8
1 / -0

An element is placed in the 2nd group and 3rd period of the periodic table, burns in presence of oxygen to form a basic oxide. The electronic configuration of the element is ____________ .

A
K, M, L (2, 8, 2)

B
K, L, M (2, 6, 4)

C
K, M, L (4, 6, 2)

D
K, L, M (2, 8, 2)

Solution

As it is in 3rd period, it has 3 shells and since it is in the second group, it contains 2 valence electrons. Thus, the electronic configuration will be 2,8,2.

Hence, it has 12 electrons, that is, magnesium.

$$2Mg + O_{2} \rightarrow 2MgO$$.

It satisfies given condition which is a basic oxide.

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

Which of the following will have maximum electron affinity?

A
$$1s^{2}2s^{2}2p^{5}$$

B
$$1s^{2}2s^{2}2p^{6}$$

C
$$1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$$

D
$$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$$

Solution
Question 10
1 / -0

Electronic configuration of sodium is :

A
$$(2,8,1)$$

B
$$(2,8,2)$$

C
$$(2,8,3)$$

D
$$(2,8,4)$$

Solution

Atomic no. of Na is 11 & thus the E.C is $$: 2, 8, 1$$

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Metal	$$K$$ $$Mg$$		$$Zn$$ $$Cu$$
Standard reduction potential (V)	$$-2.95$$ $$-2.363$$		$$-1.662$$ $$+0.340$$

Classification of Elements and Periodicity in Properties Test - 23

Result

Classification of Elements and Periodicity in Properties Test - 23

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SHARING IS CARING

Question 1

F

Cl

Br

I

Solution

Question 2

Both are rendered passive by conc. $$HNO_{3}$$

Carbides of both give methane on hydrolysis

Both give hydroxides which are basic

Both give covalent chlorides

Solution

Question 3

$$Ca$$

$$Be$$

$$Sr$$

$$Ba$$

Solution

Question 4

$$K > Mg > Zn > Cu$$

$$K > Mg > Cu > Zn$$

$$Cu > Zn > K > Mg$$

$$Mg > K > Cu > Zn$$

Solution

Question 5

$$CO,NO^{+} ,CN^{-} ,C^{2-}_{2}$$

$$N_{2} ,C_{2}^{2-} ,CO, NO$$

$$NO,CN , N_{2} ,O_{2}^{-}$$

$$NO^{+} ,C^{2-}_{2} ,O_{2}^{-} ,CO$$

Solution

Question 6

X is a metal

Y is a metal

Z is a non-metal

Y is a non-metal and Z is a metal

Solution

Question 7

$$Cl < F < S < O$$

$$O < S < F < Cl$$

$$S < O < Cl < F$$

$$F < Cl < O < S$$

Question 8

K, M, L (2, 8, 2)

K, L, M (2, 6, 4)

K, M, L (4, 6, 2)

K, L, M (2, 8, 2)

Solution

Question 9

$$1s^{2}2s^{2}2p^{5}$$

$$1s^{2}2s^{2}2p^{6}$$

$$1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$$

$$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$$

Solution

Question 10

$$(2,8,1)$$

$$(2,8,2)$$

$$(2,8,3)$$

$$(2,8,4)$$

Solution

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