Structure of Atom Test

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Structure of Atom Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The penultimate and outermost orbit of an element contains $$10$$ and $$2$$ electrons respectively. If the outermost orbit is fourth orbit, the atomic number of the element should be

A
$$12$$

B
$$22$$

C
$$32$$

D
$$40$$

Solution

Penultimate orbit i.e. the second last orbit of the atom has 10 electrons and the last orbit which is the 4th orbit has 2 electrons which means the first two orbits K-shell and L-shell are completely filled with 2 and 8 electrons respectively.

Hence, total no. of electrons or the atomic no. of the element is equal to (2+8+10+2) = 22.
Question 2
1 / -0

Magnesium hydroxide is the white milky substance in milk of magnesia. What mass of $$Mg(OH)_{2}$$ is formed when $$15 ml$$ of $$ 0.2 M - NaOH $$ is combined with $$12$$ ml of $$ 0.15 M - MgCl_{2}?$$

A
$$0.087 g$$

B
$$0.079 g$$

C
$$0.1044 g$$

D
$$0.522 g$$

Solution

$$ NaOH :- 15 ml\, 0.2 M$$
$$ MgCl_{2} :- 12 ml \, 0.15 M $$
milliequivalent of $$ NaOH = 0.2 \times 15 = 3m.e.$$
millequivalent of $$ MgCl_{2} = 0.15 \times 2 \times 12 = 3.6 m.e.$$
So only 3 m.e. of NaOH and $$ MgCl_{2}$$ reacted to give
3 m.e. of $$Mg(OH)_{2}$$
Now,
Eq.ut of $$ Mg(OH)_{2} = \dfrac{58}{2} == 29 gm$$
$$ \therefore \dfrac{ut. of Mg(OH)_{2}}{29} \times 1000 = 3 $$
$$ \therefore $$ ut. of $$ Mg(OH)_{2} = 0.087 g $$
Question 3
1 / -0

$$K$$ and $$L$$ shell of an element are completely filled and there are $$16$$ electron in $$M-$$ shell and $$2-$$ electrons in $$N-$$ shell. The atomic number of the element is

A
$$18$$

B
$$28$$

C
$$22$$

D
$$26$$

Solution

Since K and L shells are completely filled, the number of electrons in the first orbit or K-shell will be equal to 2 and the second orbit or L-shell will be 8. The M-shell has 16 electrons and N-shell has 2 electrons.

So, total no. of electrons or the atomic no. of the element = (2+8+16+2) = 28.
Question 4
1 / -0

What is the value of n for L-shell?

A
3

B
2

C
1

D
4

Solution

The Principle quantum number is denoted by $$n$$, which determines the energy of an electron.
The energy levels corresponding to numbers $$1,2,3,4,etc$$ are designated as $$K,L,M,N,etc$$.
Therefore value of $$n$$ for $$L-shell$$ is $$2$$.

Hence, the correct option is (B).
Question 5
1 / -0

Wireless mouse works on which technology?

A
IR

B
Radio frequency

C
X Rays

D
Gamma Rays

Solution

Wireless mouse works on radio frequencies.
Question 6
1 / -0

How many unpaired electrons are present in ground state of chromium $$(Z=24)$$?

A
$$1$$

B
$$5$$

C
$$6$$

D
$$0$$

Solution

Six.
The electronic configuration for Cr (24) is : 1s$$^2$$ 2s$$^2$$ 2p$$^6$$ 3s$$^2$$ 3p$$^6$$ 4s$$^1$$ 3d$$^5$$.

The no. of unpaired electrons is 6 due to half-filled subshells (4s and 3d) which are more stable than incompletely filled subshells.
Question 7
1 / -0

In the sixth period, the orbitals are filled as :

A
6s 5f 6d 6p

B
6s 4f 5d 6p

C
5s 5p 5d 6p

D
6s 6p 6d 6f

Solution
Question 8
1 / -0

In the sixth period of the extended from of periodic table, the orbitals are filled as:

A
6s, 5f, 6d, 6p

B
5s, 5p, 5d, 6p

C
6s, 6p, 6d, 6f

D
6s, 4f, 5d, 6p

Solution

In the sixth period (n=6) of the periodic table, 6s,4f, 6d, 6p orbitals are filled in the increasing order of the energy. A total of 16 orbitals are available each of which contains a maximum of 2 electrons. Thus, a total of 32 electrons can be accommodated.
Question 9
1 / -0

The effective atomic number of cobalt in the complex $$[Co(NH_3)_6]^{3+}$$ is:

A
$$36$$

B
$$33$$

C
$$24$$

D
$$30$$

Solution
Question 10
1 / -0

An element X belongs to fourth period and fifteenth group of the periodic table. Which one of the following is true regarding the outer electronic configuration of X? It has:

A
partially filled d-orbitals and completely filled s-orbital

B
completely filled s-orbital and completely filled p-orbitals

C
completely filled s- orbital and half filled p-orbitals

D
half filled d-orbitals and completely filled s-orbital

E
completely filled s-, p- and d-orbitals

Solution

Answer C.
Since, element 'X' belongs to the fourth period, n=4. For the fifteenth group, the general outer electronic configuration is $$ns^2np^3$$.

The electronic configuration of X can be written as $$X=1s^22s^22p^63s^23p^64s^23d^{10}4p^3$$. So, element 'X' has completely filled s- and d-orbitals and half-filled p-orbitals.

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Re-Attempt Weekly Quiz Competition

Structure of Atom Test - 63

Result

Structure of Atom Test - 63

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Rank

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SHARING IS CARING

Question 1

$$12$$

$$22$$

$$32$$

$$40$$

Solution

Question 2

$$0.087 g$$

$$0.079 g$$

$$0.1044 g$$

$$0.522 g$$

Solution

Question 3

$$18$$

$$28$$

$$22$$

$$26$$

Solution

Question 4

3

2

1

4

Solution

Question 5

IR

Radio frequency

X Rays

Gamma Rays

Solution

Question 6

$$1$$

$$5$$

$$6$$

$$0$$

Solution

Question 7

6s 5f 6d 6p

6s 4f 5d 6p

5s 5p 5d 6p

6s 6p 6d 6f

Solution

Question 8

6s, 5f, 6d, 6p

5s, 5p, 5d, 6p

6s, 6p, 6d, 6f

6s, 4f, 5d, 6p

Solution

Question 9

$$36$$

$$33$$

$$24$$

$$30$$

Solution

Question 10

partially filled d-orbitals and completely filled s-orbital

completely filled s-orbital and completely filled p-orbitals

completely filled s- orbital and half filled p-orbitals

half filled d-orbitals and completely filled s-orbital

completely filled s-, p- and d-orbitals

Solution

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