Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following sets of elements has strongest tendency to form anions ?

A
$$As, Se, Br$$

B
$$P, S, Cl$$

C
$$N, O, F$$

D
$$Sb, Te , I$$

Solution

Electronegativity is the tendency to attract a shared pair of electrons and this property increases moving across from left to right in a period and decreases on moving down the group. So $$N, O, F$$ are highly electronegative in nature.
Question 2
1 / -0

Which amongst the following is not an alkaline earth metal?

A
Mg

B
Ba

C
Fr

D
Sr

Solution

Alkaline earth metals are group II elements like calcium, magnesium and strontium while francium belongs to group I alkali metals with atomic number 87.
Question 3
1 / -0

The electron affinities of which group of elements are positive?

A
VIA

B
IVA

C
IA

D
Zero

Solution

Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.Electron affinity increases upward for the groups and from left to right across periods of a periodic table because the electrons added to energy levels become closer to the nucleus, thus a stronger attraction between the nucleus and its electrons. Remember that greater the distance, the less of an attraction; thus, less energy is released when an electron is added to the outside orbital. In addition, the more valence electrons an element has, the more likely it is to gain electrons to form a stable octet. The less valence electrons an atom has, the least likely it will gain electrons.ThusThe electron affinities of zero group of elements are positive.
Question 4
1 / -0

Electronic configuration of vanadium is:

A
$$[Ar] 3d^{10} 4s^1$$

B
$$[Ar] 3d^{9} 4s^2$$

C
$$[Ar] 3d^{5} 4s^1$$

D
$$[Ar] 3d^{3} 4s^2$$

Solution

Electronic configuration is the number of electrons distributed per shell.
Example:- Electronic configuration of Vanadium $$[Ar]3d^34s^2$$
Electronic configuration of Scandium $$[Ar]3d^14s^2$$.
Question 5
1 / -0

A particular atom has the 4th shell as its valence shell. If the difference between the number of electrons between K and N shell and L and M shell is zero, find the electronic configuration of its stable ion.

A
2, 8, 8, 2

B
2, 8, 18, 8

C
2, 8, 8

D
2, 8

Solution

Number of electrons in shell is given by $$2n^2$$
where $$n$$= energy level
Maximum number of electrons in $$K$$ shell
$$\longrightarrow 2n^2=2\times (1)^2=2$$ $$[n=1]$$
Maximum number of electrons in $$L$$ shell
$$\longrightarrow 2n^2=2\times (2)^2=2\times 4=8$$ $$[n=2]$$
Maximum number of electrons in $$M$$ shell
$$\longrightarrow 2n^2=2\times (3)^2=2\times 9=18$$ $$[n=3]$$
Maximum number of electrons in $$N$$ shell
$$\longrightarrow 2n^2=2\times (4)^2=2\times 16=32$$ $$[n=4]$$
Difference between $$K$$ and $$N$$ shell is zero. $$K$$ can accommodate maximum $$2$$ electron. Therefore, for difference to be zero $$N$$ should also have $$2$$ electrons.
$$\therefore K=2$$ electrons
$$N=2$$ electrons
Difference between $$L$$ and $$M$$ is also zero. Maximum electron in $$L=8$$. So, for difference to be zero $$M$$ should also have $$8$$ electron in it.
$$\therefore L=8e^-$$
$$M=8e^-$$
Electrons per shell in
$$K$$ $$L$$ $$M$$ $$N$$ [$$N$$ is the $$4^{th}$$ valence shell]
$$2$$ $$8$$ $$8$$ $$2$$
Stable ion is formed by loss of $$2$$ electrons from $$N$$ shell.
$$\therefore$$ Electronic configuration will be $$2,8,8$$ .
Question 6
1 / -0

From the set of elements given below, identify the correct order of electron affinities of the elements.
(a) $$Be > Mg > Ca > Sr > Ba$$
(b) $$F > Cl > Br > I$$
(c) $$Be > B > C > N > O$$
(d) $$Cl > F > Br > I$$

A
a and b

B
b and c

C
a and d

D
c and d

Solution

Electron affinity is the ability of an atom to accept an electron. The trend in periodic table can be written as;
(i) Electron affinity increases from left to right within a period. This is caused by the decrease in atomic radius.
(ii) Electron affinity decreases from top to bottom within a group. This is caused by the increase in atomic radius.
Also fluorine, though higher 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.
Thus options (a) and (d) are correct.
Question 7
1 / -0

Element $$A$$ has electronic configuration 2,7. $$B$$ has configuration 2,8,6. $$C$$ has configuration 2,8 while $$D$$ has 2,8,7. Which element will show similar chemical properties?

A
$$A$$ and $$C$$

B
$$A$$ and $$D$$

C
$$B$$ and $$C$$

D
$$B$$ and $$D$$
Question 8
1 / -0

The electronic configuration of the element with Atomic number 19 is___________.

A
2,8,7

B
2,9,8

C
2,8,8,1

D
2,10,7

Solution

The electronic configuration of the element with Atomic number 19 is 2,8,8,1 or $$1s^2 2s^22p^6 3s^23p^6 4s^1$$ or $$[Ar] 4s^1$$. The element is potassium. It is an alkali metal with one valence electron.
Question 9
1 / -0

The atomic number of an element 'y' is 20. The electronic configuration of the ion having inert gas configuration is:

A
2, 8, 10

B
2, 18

C
2, 10, 8

D
2, 8, 8

Solution

Electronic configuration is the number of electrons present per shell.
Inert gases have fully filled valence shell and does not react with other atoms to form molecule. $$2,8,8$$ is electronic configuration of inert gas which shows that these gases do not participate in reactions with other elements.
Example:-
(i) Electronic configuration of Argon
$$Ar(2,8,8)$$
(ii) Electronic configuration of Neon
$$Ne(2,8)$$
This shows that inert gases are quite stable.
Question 10
1 / -0

The element with electronic configuration 2, 8,6 is:

A
metallic with valency 2

B
non-metallic with valency -2

C
non-metallic with valency 2

D
none of these

Solution

The element with electronic configuration 2, 8, 6 is: Sulphur-16 which is non-metal and valency 2.

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Classification of Elements and Periodicity in Properties Test - 39

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Classification of Elements and Periodicity in Properties Test - 39

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Question 1

$$As, Se, Br$$

$$P, S, Cl$$

$$N, O, F$$

$$Sb, Te , I$$

Solution

Question 2

Mg

Ba

Fr

Sr

Solution

Question 3

VIA

IVA

IA

Zero

Solution

Question 4

$$[Ar] 3d^{10} 4s^1$$

$$[Ar] 3d^{9} 4s^2$$

$$[Ar] 3d^{5} 4s^1$$

$$[Ar] 3d^{3} 4s^2$$

Solution

Question 5

2, 8, 8, 2

2, 8, 18, 8

2, 8, 8

2, 8

Solution

Question 6

a and b

b and c

a and d

c and d

Solution

Question 7

$$A$$ and $$C$$

$$A$$ and $$D$$

$$B$$ and $$C$$

$$B$$ and $$D$$

Question 8

2,8,7

2,9,8

2,8,8,1

2,10,7

Solution

Question 9

2, 8, 10

2, 18

2, 10, 8

2, 8, 8

Solution

Question 10

metallic with valency 2

non-metallic with valency -2

non-metallic with valency 2

none of these

Solution

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