Structure of Atom Test

Result

Structure of Atom Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

If Aufbau principla is not followed the $$_{19}^{59}K$$ will be placed in:

A
$$d$$

B
$$f$$

C
$$s$$

D
$$p$$

Solution
Question 2
1 / -0

When a hydrogen atom is raised from the ground state to an excited state.

A
Both K.E. and P.E. increase

B
Both K.E. and P.E. decrease

C
The P.E. decreases and K.E. increases

D
The P.E. increases and K.E. decreases

Solution
Question 3
1 / -0

An electron with kinetic energy E collides with a hydrogen atom in the ground state. The collision will be elastic.

A
for all values of E.

B
for $$E < 10.2$$ eV.

C
for $$10.2$$ eV $$<$$ E $$< 13.6$$ eV only.

D
for $$0 <$$ E $$< 3.4$$ eV only.

Solution

$$\begin{array}{l}\text { Energy }=E \\\text { Energy at ground} \text { State for hydrogen like atom}=-13.6\frac{z^{2}}{n^{2}}=V \\\text { }\text {For hydrogen atom, } z=1 \\n=1\end{array}$$

$$E_{1}=-13.6 \times \frac{1^2}{1^{2}}=-13.6 \mathrm{eV}$$

$$\text{for 2nd shell}$$
$$\begin{aligned}E &=-13.6 \times\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) \\&=-10.2 \mathrm{eV}\end{aligned}$$
$$10.2<E<13.6$$

$$\therefore \text{Opion C is correct}$$
Question 4
1 / -0

The ratio of the wavelength of photon corresponding to the $$\alpha-$$line of Lyman series in $$H-$$atom and $$\beta-$$line of Balmer series in $$He^{+}$$ is:

A
$$1 : 1$$

B
$$1 : 2$$

C
$$1 : 4$$

D
$$3 : 16$$

Solution

$$\begin{array}{l} \cfrac { { { \lambda _{ H } } } }{ { { \lambda _{ He } } } } =\cfrac { { { R_{ H } }{ z^{ 2 } }\left( { \cfrac { 1 }{ { { 2^{ 2 } } } } -\cfrac { 1 }{ { { 4^{ 2 } } } } } \right) } }{ { { R_{ H } }\left( { \cfrac { 1 }{ { { 1^{ 2 } } } } -\cfrac { 1 }{ { { 2^{ 2 } } } } } \right) } } \Rightarrow \cfrac { { 4\left( { \cfrac { { 4-1 } }{ { 16 } } } \right) } }{ { \left( { \cfrac { { 4-1 } }{ 4 } } \right) } } =\cfrac { 1 }{ 1 } \\ { \lambda _{ H } }:{ \lambda _{ He } } \\ 1:1 \end{array}$$

Hence, the correct option is $$\text{A}$$
Question 5
1 / -0

Which of the following orbitals are nearest to nucleus?

A
4f

B
5d

C
6s

D
7p

Solution

1. 4f is closer to nucleous and not 6s.

2. But still 6s fills first because aufbau rule states that electrons fill first in orbitals with lower energy (i.e. lower value of sum of principal quantum number and azimuthal quantum number).

3. Don't confuse with energy, the one with lower principal quantum number will be inner.
Question 6
1 / -0

An element with atomic number 117 is known as:

A
nihonium

B
flerovium

C
tennessine

D
roentgenium

Solution

An element with atomic number 117 is known as tennessine. It is also called Ununseptium and has symbol Uus. Roentgenium has atomic numbe 111 and symbol Uuu/Rg.
Question 7
1 / -0

Meitnerium is IUPAC official name of an element with atomic number?

A
113

B
118

C
104

D
109

Solution

Atomic number
Name
Symbol
104 Rutherfordium
Rf
109 Meitnerium Mt
113 Ununtrium Uut
118 Ununoctium Uuo
Question 8
1 / -0

What is the atomic number of the element with symbol Uus?

A
$$117$$

B
$$116$$

C
$$115$$

D
$$114$$
Question 9
1 / -0

Which of the following statements is correct for an electron having azimuthal quantum number $$l = 2$$?

A
The electron may be in the lowest energy shell

B
The electron is in a spherical orbital

C
The electron must have spin $$m_{s} = + \dfrac {1}{2}$$

D
None

Solution

Azimuthal quantum number is also called Angular momentum quantum number. It is represented by letter $$l$$. This represents the sub-shell to which the electron belongs. It defines the shape of the orbital occupied by the electron and the angular momentum of the electron. For $$l$$=2 corresponding magnetic quantum numbers($$m_s$$)=$$-2,-1,0,1,2$$ and shape of the orbital is dumbbell shape that corresponds to the the $$d$$ $$orbital$$ as $$E_N$$ $$ \alpha$$ $$\dfrac {1}{n^2}$$. Value of n for $$l=2$$, $$n=3$$. Therefore the energy of the subshell is lowest for $$l=2$$.
Question 10
1 / -0

If EAN of a central metal ion $${ X }^{ +2 }$$ in a complex is 34. and atomic number of $$X$$ is 28. The number of monodentate ligands present in complex are:

A
3

B
4

C
6

D
2

Solution

EAN=(Atomic Number)-(Charge)+2(No. of monodentate ligands)
$$34=28-(2)+2 \times$$(No. of monodentate ligands)
$$8=2 \times$$(No. of monodentate ligands)
$$\therefore$$No. of monodentate ligands $$=4$$

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Atomic number	Name	Symbol
104	Rutherfordium	Rf
109	Meitnerium	Mt
113	Ununtrium	Uut
118	Ununoctium	Uuo

Structure of Atom Test - 48

Result

Structure of Atom Test - 48

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Question 1

$$d$$

$$f$$

$$s$$

$$p$$

Solution

Question 2

Both K.E. and P.E. increase

Both K.E. and P.E. decrease

The P.E. decreases and K.E. increases

The P.E. increases and K.E. decreases

Solution

Question 3

for all values of E.

for $$E < 10.2$$ eV.

for $$10.2$$ eV $$<$$ E $$< 13.6$$ eV only.

for $$0 <$$ E $$< 3.4$$ eV only.

Solution

Question 4

$$1 : 1$$

$$1 : 2$$

$$1 : 4$$

$$3 : 16$$

Solution

Question 5

4f

5d

6s

7p

Solution

Question 6

nihonium

flerovium

tennessine

roentgenium

Solution

Question 7

113

118

104

109

Solution

Question 8

$$117$$

$$116$$

$$115$$

$$114$$

Question 9

The electron may be in the lowest energy shell

The electron is in a spherical orbital

The electron must have spin $$m_{s} = + \dfrac {1}{2}$$

None

Solution

Question 10

3

4

6

2

Solution

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