Classification of Elements and Periodicity in Properties Test

Result

Classification of Elements and Periodicity in Properties Test - 47

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

For AX ionic crystal to exist in bcc structure, the ratio of $$\left( \dfrac{r_{cation}}{r_{anion}} \right)$$ should be:

A
between 0.41 and 0.73

B
greater than 0.73

C
less than 0.41

D
equal to 1.0

Solution

For bcc structure, the radius ratio should be greater than $$0.73$$ because the radius ratio lies between $$0.732$$ to $$1.00.$$

Hence, option B is correct.
Question 2
1 / -0

Ionisation energy of $$F^{-}$$ is $$320\ kJ\ mol^{-1}$$. The electron gain enthalpy of fluorine would be:

A
$$-320\ kJ\ mol^{-1}$$

B
$$-160\ kJ\ mol^{-1}$$

C
$$+320\ kJ\ mol^{-1}$$

D
$$+160\ kJ\ mol^{-1}$$

Solution

Electron gain enthalpy is the opposite of ionisation energy.

$$\therefore Electron \, Gain\, Enthalphy=-IE$$

$$\therefore EGE \ (F^-)=-320\ kJ/mol$$
Question 3
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
$$F < Cl < O < S$$

Solution

Correct option is $$B:O<S<F<Cl$$
$$F$$ and $$Cl$$ have high tendency to gain an $$e^-$$ and stabilize themselves by fulfilling outermost shell. $$F$$ being small in size & more electronegative than $$Cl$$, $$Cl$$ accepts $$e^-$$ more readily than $$F$$. On the other hand $$O$$ with more electronegativity than $$S$$ and is small in size has less $$e^-$$ gaining tendency than $$O$$. Hence the order $$O<S<F<Cl$$.
Question 4
1 / -0

Flourine is the best oxidising agent because it has:

A
highest electron affinity

B
highest $$E_{red}^{\circ}$$

C
highest $$E_{oxid}^{\circ}$$

D
lowest electron affinity

Solution

Oxidizing agent means element, which oxidizes others by itself getting reduced.
Oxidation occurs by loss of $$e^-$$
For eg:-
$$\overset {0} H\longrightarrow \overset {+1} {H^+}+e^-$$
loss of $$e^-$$
$$ {F+e^-\longrightarrow F^-}$$
Here $$F$$ oxidizes $$H$$ and itself gets reduced.
(Refer to Image)
$$\because F$$ has high tendency to gain $$e^-$$ and to stabilize itself, i.e. have high $$EGE$$ or $$e^-$$ affinity. Hence accepts $$e^-$$ from $$H$$ and oxidized $$H$$ and act as good oxidizing agent.
Question 5
1 / -0

Select the group where $$EN$$ increases down the group.

A
$$F, Cl, Br$$

B
$$Li, Na, K$$

C
$$Ca, Sr, Ba$$

D
$$Zn, Cd, Hg$$

Solution

Generally, electronegativity increases as we move from left to right across a period and decreases as we move down a group. But in the case of Zn, Cd and Hg which belong to 3d, 4d and 5d series of transition elements (group 12), EN increases down the group. This is due to the imperfect screening effect of d and f-orbitals in transition elements.

Hence, option D is correct.
Question 6
1 / -0

Which of the following electronic configurations corresponds to elements with largest negative electron gain enthalpy?

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

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

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

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

Solution

Electron gain enthalpy is the amount of energy released when an electron is added to a isolated gaseous atom.
Halogens have greater electron gain enthalpy values. So, $${ 1S }^{ 2 }{ 2S }^{ 2 }{ 2P }^{ 6 }{ 3S }^{ 2 }{ 3P }^{ 5 }$$ has greater value.
Question 7
1 / -0

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 the atom. Hence 1st electron affinity $$\left( E{ A }_{ 1 } \right) $$ is always taken as negative. The addition of second electron to the negativity charged ions is opposed by Coulombic repulsion, hence energy has to be supplied for the addition of second electron. Hence $$\left( E{ A }_{ 2 } \right) $$ is taken as positive.

Addition of an electron to $${F}^{-}$$ and $${O}^{-}$$ ions will involve repulsions, hence their $$\left( E{ A }_{ 2 } \right) $$ will be positive.
Due to stable fully filled configuration of $${Na}^{+}$$ $$\left( 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 6 } \right) $$ it has lower value of electron affinity than $$O$$. Hence $$O$$ has highest electron affinity.
Question 8
1 / -0

If periodic table would have contained 10 period, maximum elements in that periodic table would be:

A
72

B
190

C
144

D
290

Solution

For, even no. of period $$=\cfrac {(n+2)^2}{2}$$

So, $$10$$ periods $$=\cfrac {(10+2)^2}{2}$$
$$=\cfrac {144}{2}=72$$

$$\therefore$$ Number of elements $$=72$$
Question 9
1 / -0

The chemical activity of an atom is dependent on the number of __________.

A
protons

B
neutrons

C
orbits

D
valence electrons

Solution

The chemical activity of an atom is dependent on the number of valence electrons.
An atom with octet configuration does not participate in chemical reactions and is known as chemically inert.
Question 10
1 / -0

Few general names are given along with their valence shell configurations. Mark the incorrect name.

A
$$ns^{2} np^{6}$$ - Noble gases

B
$$ns^{2} np^{5}$$ - Halogens

C
$$ns^{1}$$ - Alkali metals

D
$$ns^{2} np^{2}$$ - Chalcogens

Solution

Chalcogens are group 16 elements with valence shell electronic configuration $$ns^2np^4$$. They belong to oxygen family. Group 14 element (Carbon family) have valence shell electronic configuration $$ns^2np^2$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Classification of Elements and Periodicity in Properties Test - 47

Result

Classification of Elements and Periodicity in Properties Test - 47

Score

-

Rank

-

SHARING IS CARING

Question 1

between 0.41 and 0.73

greater than 0.73

less than 0.41

equal to 1.0

Solution

Question 2

$$-320\ kJ\ mol^{-1}$$

$$-160\ kJ\ mol^{-1}$$

$$+320\ kJ\ mol^{-1}$$

$$+160\ kJ\ mol^{-1}$$

Solution

Question 3

$$Cl < F < S < O$$

$$O < S < F < Cl$$

$$S < O < Cl < F$$

$$F < Cl < O < S$$

Solution

Question 4

highest electron affinity

highest $$E_{red}^{\circ}$$

highest $$E_{oxid}^{\circ}$$

lowest electron affinity

Solution

Question 5

$$F, Cl, Br$$

$$Li, Na, K$$

$$Ca, Sr, Ba$$

$$Zn, Cd, Hg$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$${F}$$

$$O$$

$${O}^{-}$$

$${Na}^{+}$$

Solution

Question 8

72

190

144

290

Solution

Question 9

protons

neutrons

orbits

valence electrons

Solution

Question 10

$$ns^{2} np^{6}$$ - Noble gases

$$ns^{2} np^{5}$$ - Halogens

$$ns^{1}$$ - Alkali metals

$$ns^{2} np^{2}$$ - Chalcogens

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.