Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 61

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Weekly Quiz Competition

Question 1
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Which of the following species has the highest electron affinity ?

A
$$F^{-} $$

B
$$O$$

C
$$O^{-}$$

D
$$Na^{+} $$

Solution

Energy is released when an electron is added to $$O$$ but energy has to be spent when an electron is added to O−,F−.
Question 2
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The order of first electron affinity of $$O, S$$ and $$Se$$ is:

A
$$O > S > Se$$

B
$$S > Se > O$$

C
$$Se > O > S$$

D
$$S > O > Se$$

Solution

O has an exceptionally smaller value of electron affinity $$(-141\ kJmol^{-1})$$ due to smaller atomic size than sulphur (weaker electron-electron repulsion in 3p-subshell). It is less than Se and Te also.

$$ S>Se>O$$

Thus, Option B is correct.
Question 3
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There are four elements p, q, r and s having atomic numbers $$Z-1$$, $$Z$$, $$Z+1$$ and $$Z+2$$ respectively. If the element q is an inert gas, select the correct statements:
(i) p has most negative electron gain enthalpy in the respective period.
(ii) r is an alkali metal.
(iii) s exists in +2 oxidation state.

A
(i) and (ii) only

B
(ii) and (iii) only

C
(i) and (iii) only

D
(i), (ii) and (iii)

Solution

As 'q' is a noble gas, p, r and s having atomic number $$Z-1$$, $$Z+1$$ and $$Z+2$$ should belong to halogen, alkali metal and alkaline earth metal respectively.

As halogen has one electron less than stable noble gas configuration, it has a greater tendency to accept an additional electron forming an anion.

Alkaline earth metal having valence shell configuration $$ns^{2}$$ exists in $$+2$$ oxidation state.
Question 4
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Statement-1 : Electron affinity of fluorine is less than that of chlorine.
Statement-2 : Chlorine can accommodate an electron by utilising its vacant 3d-orbital.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True

Solution

The electron affinity of fluorine is less than chlorine because fluroine is exceptionally small in size because of which energy has to be supplied to introduce an extra electron to overcome the electronic repulsions between the incoming electron and the electrons already present.
Question 5
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The species among the following having the highest electron affinity is:

A
$$O$$

B
$$F^{-}$$

C
$$Cl^{-}$$

D
$$O^{-}$$

Solution

Fluoride and chloride ions have a complete octet configuration. Hence, they are not ready to accept an electron. When adding one more electron to the uninegative oxygen anion, electron-electron repulsion will occur and makes the reaction endothermic.

Hence, the oxygen atom has the highest electron affinity among other given options.
Question 6
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The electron gain enthalpies of halogens in kJ mol$$^{-1}$$ are as given as F = -332, Cl = -349, Br = -324, I = -295. The less negative value for F as compared to that of Cl is due to :

A
strong electron - electron repulsions in the compact 2p-sub shell of F

B
weak electron - electron repulsions in the bigger 3p-sub shell of Cl

C
smaller electronegativity value of F than Cl

D
both (A) and (B)

Solution

Due to small size of F atom, the electron-electron repulsions in compact 2-p subshell are large and hence the incoming electron is not accepted with the same ease as is the case with Cl (less electron - electron repulsions).
Question 7
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The correct order of electron affinity is :

A
$$\displaystyle Be< B< C< N$$

B
$$\displaystyle Be< N< B< C$$

C
$$\displaystyle N< Be< C< B$$

D
$$\displaystyle N< C< B< Be$$

Solution

The increasing order of electron affinity is $$\displaystyle Be < N < B < C $$.
In a period, electron affinity becomes more negative as one goes from left to right.
The electron affinity of N is positive due to extra stability due to half filled p orbital.
Question 8
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Directions For Questions

The periodicity is related to the electronic configuration. That is, all chemical and physical properties are a manifestation of the electronic configurations of the elements.
The atomic and ionic radii generally decrease in a period from left to right. As a consequence, the ionization enthalpies generally increase the electron gain enthalpies become more negative across a period. In other words, the ionization enthalpy of the extreme left element in a period is the least and the electron gain enthalpy of the element on the extreme right is the highest negative. This results into high chemical reactivity at the two extremes and the lowest in the center. Similarly down the group, the increase in atomic and ionic radii result in a gradual decrease in ionization enthalpies and a regular decrease (with an exception in some third-period elements) in electron gain enthalpies in the case of main group elements.
These properties can be related to the:
(i) Reducing and oxidizing behaviour of the elements.
(ii) The metallic and non-metallic character of an element.
(iii) The acidic, basic, amphoteric and neutral character of the oxides of the elements.

...view full instructions

Among $$Al_2O_3, SiO_2, P_2O_3$$ and $$SO_2$$ the correct order of acid strength is:

A
$$Al_2O_3 < SiO_2 < SO_2 < P_2O_3$$

B
$$SiO_2 < SO_2 < Al_2O_3 < P_2O_3$$

C
$$SO_2 < P_2O_3 < SiO_2 < Al_2O_3$$

D
$$Al_2O_3 < SiO_2 < P_2O_3 < SO_2$$

Solution

As electronegativity difference between element and oxygen decreases the acidic character of oxides increases. The electronegativity also increases with increasing oxidiation state. In general, as non-metallic character increases across the period, the acidic character of their oxides increases.
Thus the correct order of acidic strength can be given as:
$$Al_2O_3<SiO_2<P_2O_3<SO_2$$
Question 9
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Which of the following order(s) for electron affinity is(are) correct?

(a)$$\displaystyle S> O< Se$$
(b)$$\displaystyle CI> F$$
(c)$$\displaystyle S> O$$
(d)$$\displaystyle O> S$$
(e)$$\displaystyle N> P$$
(f)$$\displaystyle C> N$$

A
$$a,b,c,e$$

B
$$a,b,c,f$$

C
$$b,c,d,e$$

D
$$b,c,f$$

Solution

In a period, with the increase in atomic number, electron affinity becomes more negative due to the increase in the effective nuclear charge. In a group, it becomes less negative on moving from top to bottom.

N has positive electron affinity due to the extra stability of exactly half-filled p orbitals.
In case of $$\displaystyle F^-$$ ion and $$\displaystyle O^{6-} $$ ion, due to small size, the inter-electronic repulsion is higher. Hence, less energy is evolved in the formation of the ion.

Due to this, the orders (b) CI > F and (c) S > O is correct.

The orders a, b, c, and f are correct whereas the orders d and e are incorrect.

Hence, the correct option is $$\text{B}$$
Question 10
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The correct sequence of the electron affinity of $$C, N, O$$ and $$F$$ is:

A
$$\displaystyle C> N< O< F$$

B
$$\displaystyle O> N> C> F$$

C
$$\displaystyle C< N> O< F$$

D
$$\displaystyle C> N> O> F$$

Solution

The correct sequence of the electron affinity of $$C, N, O$$ and $$F$$ is $$\displaystyle C> N< O< F$$. The electron affinity of $$C$$ is greater than that of $$N$$. The electronic configurations of $$C$$ and $$N$$ are $$1s^22s^22p^2$$ and $$1s^22s^22p^3$$ respectively. In case of $$N$$, the $$2p$$ orbital is half filled and stable.

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Classification of Elements and Periodicity in Properties Test - 61

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Classification of Elements and Periodicity in Properties Test - 61

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SHARING IS CARING

Question 1

$$F^{-} $$

$$O$$

$$O^{-}$$

$$Na^{+} $$

Solution

Question 2

$$O > S > Se$$

$$S > Se > O$$

$$Se > O > S$$

$$S > O > Se$$

Solution

Question 3

(i) and (ii) only

(ii) and (iii) only

(i) and (iii) only

(i), (ii) and (iii)

Solution

Question 4

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

Solution

Question 5

$$O$$

$$F^{-}$$

$$Cl^{-}$$

$$O^{-}$$

Solution

Question 6

strong electron - electron repulsions in the compact 2p-sub shell of F

weak electron - electron repulsions in the bigger 3p-sub shell of Cl

smaller electronegativity value of F than Cl

both (A) and (B)

Solution

Question 7

$$\displaystyle Be< B< C< N$$

$$\displaystyle Be< N< B< C$$

$$\displaystyle N< Be< C< B$$

$$\displaystyle N< C< B< Be$$

Solution

Question 8

$$Al_2O_3 < SiO_2 < SO_2 < P_2O_3$$

$$SiO_2 < SO_2 < Al_2O_3 < P_2O_3$$

$$SO_2 < P_2O_3 < SiO_2 < Al_2O_3$$

$$Al_2O_3 < SiO_2 < P_2O_3 < SO_2$$

Solution

Question 9

$$a,b,c,e$$

$$a,b,c,f$$

$$b,c,d,e$$

$$b,c,f$$

Solution

Question 10

$$\displaystyle C> N< O< F$$

$$\displaystyle O> N> C> F$$

$$\displaystyle C< N> O< F$$

$$\displaystyle C> N> O> F$$

Solution

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