Complete the table given below by identifying P , Q , R and S . Element Symbol No. of Protons No. of Neutrons No. of Electrons Sodium Na 23 11 11 P 11 Chlorine Cl 35 17 Q 18 17 Uranium R 92 146 92 S F 19 9 9 10 9

Element Symbol No. of Protons No. of Neutrons No. of Electrons Sodium Na 23 11 11 12 11 Chlorine Cl 35 17 17 18 17 Uranium U 238 92 92 146 92 Fluorine F 19 9 9 10 9 As we know that A = Z + n , so n = A - Z , A = 23 , Z = 11 so n = 23 - 11 = 12 , Number of neutrons= 12 so p = 12 , Number of protons is equal to the atomic number so Q = 17 . Mass number is Z + n , where Z = Atomic number = number of protons . So R = 92 + 146 = 238 . S = Flourine , because atomic number of fluorine is 9 .

Draw orbital diagram of Ca 2 + 20 40 ion and state the number of three fundamental particles present in it.

Orbital diagram of Ca 2 + 20 40 ion is as follows: In Ca 2 + 20 40 where + 2 indicate that calcium lost its two valence electrons. So, its fundamental particles are: Number of electrons = 18 Number of protons = 20 Number of neutrons = 20

Write down the electronic configuration of the following: X 27 13 Write down the number of electrons in X .

X 27 13 Given symbol shows that: Atomic number Z = 13 . Electronic configuration = 2 , 8 , 3 So, number of electrons = 13 , The atomic number = 13 is of aluminium, it is a lightweight silvery white metal of main group 13 .

Write down the electronic configuration of the following: Y 35 17 Write down the number of neutrons in Y .

Y 35 17 Given symbol represent that: Atomic number = 17 . Electronic configuration 2 , 8 , 7 Hence, number of neutrons 35 - 17 = 18 . From the electronic configuration we can conclude that the given element is chlorine, as atomic number of chlorine is 17 . Chlorine Cl exists as a yellow green gas with a pungent smell. It has the atomic number 17 in the periodic table and belongs to Group 17 , the halogens. It is a non-metal with the symbol Cl .

E M B I B E

Concise Chemistry Class 9>Chapter 4 - Atomic Structure and Chemical Bonding>EXERCISE 4(B)

S P Singh Solutions for Chapter: Atomic Structure and Chemical Bonding, Exercise 2: EXERCISE 4(B)

Author:S P Singh

S P Singh Chemistry Solutions for Exercise - S P Singh Solutions for Chapter: Atomic Structure and Chemical Bonding, Exercise 2: EXERCISE 4(B)

Attempt the practice questions on Chapter 4: Atomic Structure and Chemical Bonding, Exercise 2: EXERCISE 4(B) with hints and solutions to strengthen your understanding. Concise Chemistry Class 9 solutions are prepared by Experienced Embibe Experts.

Questions from S P Singh Solutions for Chapter: Atomic Structure and Chemical Bonding, Exercise 2: EXERCISE 4(B) with Hints & Solutions

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9th ICSE

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Concise Chemistry Class 9>Chapter 4 - Atomic Structure and Chemical Bonding>EXERCISE 4(B)>Q 2

Complete the table given below by identifying $P, Q, R and S$ .

Element	Symbol	No. of Protons	No. of Neutrons	No. of Electrons
Sodium	${}_{11}{Na}^{23}$	$11$	$P$	$11$
Chlorine	${}_{17}{Cl}^{35}$	$Q$	$18$	$17$
Uranium	$R$	$92$	$146$	$92$
$S$	${}_{9}F^{19}$	$9$	$10$	$9$