Structure of Atom Test

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Structure of Atom Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following is the correct order of energies of 3p, 3d, 4s and 4p orbitals as per Aufbau principle ?

A
$$3p<3d<4s <4p $$

B
$$3p < 4s <3 d<4p$$

C
$$ 3d<43<4p<3p$$

D
$$3d<3p< 4p<4$$

Solution

Correct order of energies of the given orbital as per Aufbau principle is:
$$3p < 3d < 4s < 4p$$
Question 2
1 / -0

Which electron,orbitals are designated by
(i) n= 2, l=1, m= 0;
(ii)n= 3, l= 2, m= 0;
(iii)n = 4, l = 2, m = 1 and
(iv) n = 5, I= 3, m =3 respectively ?

A
2p, 3d, 4d and 5f, respectively

B
2p, 3d, 4d and 5d, respectively

C
2p, 3d, 4f and 5f, respectively

D
2p, 3p, 4d and 5f, respectively

Solution

For $$(i) n = 2, l= 1, m =.0$$; the orbital occupied by electron is 2p.

For$$ (ii) n = 3, l= 2, m = 0$$ ;the orbital occupied by electron is 3d.

For$$ (iii) n = 4, l= 2, m =1$$; the orbital occupied by electron is 4d.

For $$(iv) n = 5, l = 3, m = 3$$; the orbital occupied by electron is 5f.
Question 3
1 / -0

The maximum number of electrons in any shell is given by____ rule.

A
2n

B
$$n^2$$

C
$$2n^2$$

D
$$4n^2$$

Solution

Maximum no of electrons in a shell $$={2n}^{2}$$
this is called $${2n}^{2} $$rule
for example,second shell $$\Rightarrow n=2$$
to $${2n}^{2}$$rule,no of electrons in $${2}^{nd}$$ shell=$$2 \left( 2 \right)^2$$=8
Question 4
1 / -0

The Ionisation potential of Hydrogen is $$2.17\times 10^{-11}erg/ atom$$. The energy of the electron in the second orbit of the hydrogen atom is?

A
$$-\dfrac {2.17\times 10^{-11}}{2}$$

B
$$-\dfrac {2.17\times 10^{-11}}{2^{2}}$$

C
$$-\dfrac {2.17\times 10^{-7}}{2^{2}}$$

D
$$-\dfrac {2.17\times 10^{11}}{2^{2}}$$

Solution

Energy of electron hydrogen atom is given by,
$$E_n=-\dfrac{2.178\times 10^{-11}}{n^2}ergatom^{-1}$$
Therefore, for second orbit i.e $$n=2$$
$$E_2=-\dfrac{2.178\times 10^{-11}}{2^2}ergatom^{-1}$$
Question 5
1 / -0

In Bohr series of lines of hydrogen spectrum, the third line from the red corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?

A
$$2\rightarrow 5$$

B
$$3\rightarrow 2$$

C
$$5\rightarrow 2$$

D
$$4\rightarrow 1$$

Solution

$$\bf{Explanation-}$$
Line Spectrum of Hydrogen like atoms are given by:
$$\dfrac{1}{\lambda} = RZ^2(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2})$$
Where R is called Rhydberg constant, $$R = 1.097 \times 10^7$$,
Z is atomic number.
$$n_1= 1,2 ,3….$$
$$n_2= n_1+1, n_1+2 ,……$$

The electron has minimum energy in the first orbit and its energy increases as $$n$$ increases. Here $$n$$ represents the number of orbits.

Red line means it is the visible spectra and visible spectra means it is Balmer series that is $$n=2$$
So, the third line after $$n=2$$ will be $$n=5$$ that is Pfund series.
Therefore, the electrons jump from $$n=5$$ to $$n=2$$.

$$\bf{Correct \ Option}-$$ Option $$C$$
Question 6
1 / -0

Which of the following electronic configuration is not possible?

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

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

C
$$\left[ Ar \right] 3{ d }^{ 10 },4{ s }^{ 2 },4{ p }^{ 2 }$$

D
$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 2 },3{ s }^{ 1 }$$
Question 7
1 / -0

The energy levels for $$_{Z}A^{(+Z - 1)}$$ can be given by:

A
$$E_{n}$$ for $$A^{(+Z - 1)} = Z^{2} \times E_{n}$$ for $$H$$

B
$$E_{n}$$ for $$A^{(+Z - 1)} = Z \times E_{n}$$ for $$H$$

C
$$E_{n}$$ for $$A^{(+Z - 1)} = \dfrac {1}{Z^{2}} \times E_{n}$$ for $$H$$

D
$$E_{n}$$ for $$A^{(+Z - 1)} = \dfrac {1}{Z} \times E_{n}$$ for $$H$$

Solution

According to Bohr's model, $$E = 13.6 \space kcal$$ for Hydrogen.
$$E = 13.6\dfrac{Z^2}{n}$$
Hence $$E_n$$ for $$A^{(+Z - 1)} =Z^2 \times E_n$$ for $$H.$$
Question 8
1 / -0

The electrons, identified by quantum numbers $$n$$ and $$l$$ (i) $$n=4,l=1$$, (ii) $$n=4,l=0$$ (iii) $$n=3,l=2$$ (iv) $$n=3,l=1$$ can be place in order of increasing energy, from the lowest to highest, as:

A
(iv)< (ii)< (iii)< (i)

B
(ii)< (iv)< (ii)< (iii)

C
(i)< (iii)< (ii)< (ii)

D
(iii)< (i)< (iv)< (ii)

Solution

(i) $$4p$$ (ii) $$4s$$ (iii) $$3d$$ (iv) $$3p$$

According to Aufbau rule, order of increasing energy is $$3p< 4s< 3d< 4p$$
Question 9
1 / -0

A certain negative ion $${X}^{2-}$$ has in its nucleus $$18$$ neutrons and $$18$$ electrons in its extra nuclear structure. What is the mass number of $$X$$?

A
$$34$$

B
$$32$$

C
$$36$$

D
$$39$$

Solution

Number of electrons in $$ \displaystyle X^{2-}$$ ion $$ \displaystyle =18$$

Number of electrons in $$ \displaystyle X $$ atom $$ \displaystyle =18-2=16$$

Mass number $$ \displaystyle =$$ number of electrons in neutral atom $$ \displaystyle +$$ number of neutrons

$$ \displaystyle =16+18=34$$
Question 10
1 / -0

A photon of energy 12.75 ev is completely absorbed by a hydrogen atom initially In ground state. The principle quantum number of the excited state is

A
1

B
3

C
4

D
$$\infty $$

Solution

Transition energy for Hydrogen atom is given by:

$$\Delta E=-R(1/n_2^2-1/n_1^2)$$

where $$R=13.6\ ev,n_1=1,n_2=n$$

$$12.75=-13.6(1/n^2-1)$$

$$-0.9375+1=1/n^2$$

$$n^2=16$$

$$n=4$$

So, the corret option is $$C$$.

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Structure of Atom Test - 47

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Structure of Atom Test - 47

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Question 1

$$3p<3d<4s <4p $$

$$3p < 4s <3 d<4p$$

$$ 3d<43<4p<3p$$

$$3d<3p< 4p<4$$

Solution

Question 2

2p, 3d, 4d and 5f, respectively

2p, 3d, 4d and 5d, respectively

2p, 3d, 4f and 5f, respectively

2p, 3p, 4d and 5f, respectively

Solution

Question 3

2n

$$n^2$$

$$2n^2$$

$$4n^2$$

Solution

Question 4

$$-\dfrac {2.17\times 10^{-11}}{2}$$

$$-\dfrac {2.17\times 10^{-11}}{2^{2}}$$

$$-\dfrac {2.17\times 10^{-7}}{2^{2}}$$

$$-\dfrac {2.17\times 10^{11}}{2^{2}}$$

Solution

Question 5

$$2\rightarrow 5$$

$$3\rightarrow 2$$

$$5\rightarrow 2$$

$$4\rightarrow 1$$

Solution

Question 6

$$1{ s }^{ 2 }2{ s }^{ 2 }$$

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

$$\left[ Ar \right] 3{ d }^{ 10 },4{ s }^{ 2 },4{ p }^{ 2 }$$

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

Question 7

$$E_{n}$$ for $$A^{(+Z - 1)} = Z^{2} \times E_{n}$$ for $$H$$

$$E_{n}$$ for $$A^{(+Z - 1)} = Z \times E_{n}$$ for $$H$$

$$E_{n}$$ for $$A^{(+Z - 1)} = \dfrac {1}{Z^{2}} \times E_{n}$$ for $$H$$

$$E_{n}$$ for $$A^{(+Z - 1)} = \dfrac {1}{Z} \times E_{n}$$ for $$H$$

Solution

Question 8

(iv)< (ii)< (iii)< (i)

(ii)< (iv)< (ii)< (iii)

(i)< (iii)< (ii)< (ii)

(iii)< (i)< (iv)< (ii)

Solution

Question 9

$$34$$

$$32$$

$$36$$

$$39$$

Solution

Question 10

1

3

4

$$\infty $$

Solution

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