Structure of Atom Test

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Structure of Atom Test - 45

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Question 1
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The electronic configuration of an atom/ion can be defined by which of the following?

A
Aufbau principle

B
Pauli's exclusion principle

C
Hund's rule of maximum multiplicity

D
All of the above

Solution

The electronic configuration of an atom/ion can be defined by the following rules
(A) Aufbau principle
In the ground state of the atoms, the orbitals are filled with electrons in order of increasing energy.

(B) Pauli's exclusion principle
Two electrons in atom cannot have the same set of all four quantum numbers.

(C) Hund's rule of maximum multiplicity
When several orbitals of equal energy are available, the electrons first fill all the orbitals singly before pairing in any of these orbitals.
Question 2
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Which among the following is incorrect of $$_{5}{B}$$ in normal state?

A

B

C

D

Solution

The option (C) is correct of $$_5B$$ in normal state. It is against Pauli exclusion principle and not Hund's rule of maximum multiplicity.
Pauli exclusion principle : Two electrons in an atom cannot have the same set of all four quantum numbers.
In the given diagram, two electrons in s subshell have same spin. This violets Pauli exclusion principle.

Hund's rule : When several orbitals of equal energy (degenerate orbitals) are available, the electrons first fill all the orbitals singly before pairing in any of these orbitals.

According to Aufbau principle, orbitals are filled in order of increasing value of $$(n+l)$$.
If two orbitals have same value of $$(n+l)$$, then the orbital with lower value of $$n$$ will be filled first.
Question 3
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After filling the $$4d$$-orbitals, an electron will enter in:

A
$$4p$$

B
$$4s$$

C
$$5p$$

D
$$4f$$

Solution

After filling the 4d-orbital, an electron will enter in 5p orbital.

The 4d sub-energy level is at a lower energy than the 5p sub-energy level
For 4d, $$\displaystyle n=4$$ and $$\displaystyle l=2$$. Hence, $$\displaystyle (n+l) = 4+2=6$$
For 5p, $$\displaystyle n=5$$ and $$\displaystyle l=1$$. Hence, $$\displaystyle (n+l) = 5+1=6$$

As the value of $$\displaystyle (n+l)$$ for 4d orbital is same as that of 5p orbital, 4d orbital is filled before 5p orbital as 4d orbital has lower value of n than 5 p orbital.

Note: If two orbitals have same value of $$\displaystyle (n+l)$$, then the orbital with lower value of n will be filled first.
Question 4
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The orbital diagram which violates the 'aufbau principle' is:

A

B

C

D

Solution

The orbital diagram in which 'aufbau principle' is violated, is represented by the option (B).

Aufbau principle:
In the ground state of the atoms, the orbitals are filled with electrons in order of increasing energy.

The 2s sub-energy level is at a lower energy than the 2p sub-energy level
For 2s, $$\displaystyle n=2$$ and $$\displaystyle l=0$$. Hence, $$\displaystyle (n+l) = 2+0=2$$
For 2p, $$\displaystyle n=2$$ and $$\displaystyle l=1$$. Hence, $$\displaystyle (n+l) = 2+1=3$$
As the value of $$\displaystyle (n+l)$$ for $$2s$$ orbital is less than that of $$2p$$ orbital, $$2s$$ orbital is filled before $$2p$$ orbital.
Question 5
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The mass of an atom of carbon is:

A
$$1\, g$$

B
$$1.99 \times 10^{-23}\, g$$

C
$$1/12 \, g$$

D
$$1.99 \times 10^{23}\, g$$

Solution

1 mole of $$C$$ weighs 12 g.

1 mole of $$C$$ contains $$ \displaystyle 6.02 \times 10^{23}$$ atoms.

$$ \displaystyle 6.02 \times 10^{23}$$ atoms of C weighs 12 g.

1 atom of C weighs $$ \displaystyle \dfrac {12} {6.02 \times 10^{23}}=1.99 \times 10^{-23}$$ g.
Question 6
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For $${Li}^{2+}$$, when an electron falls from a higher orbit to $$n$$th orbit, all the three types of lines, i.e., Lyman, Balmer and Paschen was found in the spectrum. Here, the value of '$$n$$' will be ___.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$\cfrac{1}{\lambda}={ RZ }^{ 2 }\left( \cfrac { 1 }{ { n }_{ 1 }^{ 2 } } -\cfrac { 1 }{ { n }_{ 2 }^{ 2 } } \right) $$
given,$${n}_{1}=n$$
lyman series is obsereved only if $${n}_{1}=n=1$$
Question 7
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26.8 gm of $$Na_2SO_4\ . nH_2O$$ contains 12.6 gm of water. The value of 'n' is:

A
1

B
10

C
6

D
7

Solution

$$NaSO_4 \cdot nH_2O$$
Molar mass $$= (142 +18 n)$$
Here number of moles of water$$=\dfrac{12.6}{26.8}$$
Mass of water $$= \dfrac{12.6}{26.8}\times (142 +18 \, n)$$
$$18 n =\dfrac{12.6}{26.8}\times (142 +18n)$$
$$n=7$$
Question 8
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After $$np$$-orbitals are filled, the next orbital filled will be:

A
$$\left(n+1\right)s$$

B
$$\left(n+2\right)p$$

C
$$\left(n+1\right)d$$

D
$$\left(n+2\right)s$$

Solution

After np-orbitals are filled, the next orbital filled will be $$\displaystyle (n+1)s$$.
According to Aufbau principle, orbitals are filled in order of increasing value of $$\displaystyle (n+l)$$.
If two orbitals have same value of $$\displaystyle (n+l)$$, then the orbital with lower value of n will be filled first.
For np orbital, $$\displaystyle (n+l) = (n+1)$$
For $$\displaystyle (n+1)s$$ orbital, $$\displaystyle (n+l) = (n+1+0)=(n+1)$$
For $$\displaystyle (n+2)p$$ orbital, $$\displaystyle (n+l) = (n+2+1)=(n+3)$$
For $$\displaystyle (n+1)d$$ orbital, $$\displaystyle (n+l) = (n+1+2)=(n+3)$$
For $$\displaystyle (n+2)s$$ orbital, $$\displaystyle (n+l) = (n+2+0)=(n+2)$$
Question 9
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The molar mass of $$N_2O$$ as well as $$CO_2$$ is $$44\, g\, mol^{-1}$$. At $$25^o C$$ and 1 atm pressure, 1 L $$N_2O$$ contains n molecules of gas. The number of $$CO_2$$ molecules in 2 L under same conditions will be:

A
$$n$$

B
$$2n$$

C
$$\frac{n}{2}$$

D
$$\frac{n}{4}$$

Solution

We have,
$$PV = nRT$$
P, R, T are same, hence we get
$$\dfrac{V_1}{V_2} = \dfrac{n_1}{n_2}$$
n =$$\dfrac{no. of molecules}{N_A}$$ =number of moles
$$\dfrac{1}{2} = \dfrac{n}{molecules of CO_2}$$
hence, the number of molecules of $$CO_2$$ $$= 2n$$
Question 10
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A quantity of aluminium has a mass of $$54 g$$. What is the mass of same number of magnesium atoms?

A
$$12.1 g$$

B
$$24.3 g$$

C
$$48.6 g$$

D
$$97.2 g$$

Solution

Weight of alluminium = 54 g
moles of alluminium = $$\dfrac{54}{27} = 2$$
moles of magnesium is also = 2
mass of magnesium ion becomes = $$moles * molecular weight = 2*24 = 48g$$