Question

Q1.the divalent cation of an atom contains 10 electrons .if mass number of an element is 24 find the number of electrons, protons and neutrons in the ion and atomic number of the element.
Q2. the monovalent anion of an atom has 18 electrons mass number is 35 find the number of electrons, protons and neutrons and atomic number
Q3.the divalent cation of an element contains completely filled L shell .find the number of electrons ,protons and neutrons in the cation if its mass number is 24
q4.the monovalent anion of an element completely filled L shell find atomic number, electronic configuration, electrons, protons and neutrons of mass number is 19
please tell these all 4​

Answer 1

answer:

1. electron= 10 +2 = 12(as it is di valent cation i.e. having two e less than element)

proton = 12 = no.of electron in normal atom but it's divalent cation so, it's atom has 12 e.

neutron = 24-12= mass no. - no. of proton= 12

atomic no. = no.of proton = 12

Explanation:

2. no. of e = 18 -1 = 17 ( as it is monovalent anion i.e. having one more e than element)

no. of p = no. of e = 17

no. of n = mass no. - no. of p = 35-17 = 18

atomic no. = no. of p = 17

3. for L shell no. of e = 2n^2 here for l shell n=2

= 8 e

in cation no. of e = k shell e + l shell e

= 2+8=10

so, no of e in element =12. ( divalent cation)

no. of p= 12. (( use above used formulas now))

no. of n = 24 - 12 =12

atomic no. = 12.

4. no of e in anion = 10 ( same as calculated in 3)

no of e in element = 10-1 = 9 ( monovalent anion)

no of p = 9

no of n = 19-9= 10

atomic no = 9

electronic configuration = 2,7

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Answer 2

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$\large\boxed{1}$

Given :-

• cation : (divalent)
• electrons , e = 10
• mass number , A = 24

To Find :-

no. of electrons in element

= e + valence

= 10 + 2

= 12 electrons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$no. of electrons in the ion

= 10 ( given )

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$no. of protons in the ion

= no. of electrons

= 12 protons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$no. of neutrons in the ion

= A - no. of protons

= 24 - 12

= 12 neutrons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$atomic no.

= no. of protons

= 12

$\green{\rule{5.2cm}{0.03cm}}$

$\large \boxed{2}$

Given :-

• anion : (monovalent)
• mass number , A = 35

To Find :-

$\bigstar$no. of electrons

= 18 - 1

= 17 electrons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$no. of protons

= no. of electrons

= 17 protons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$no. of neutrons

= A - no. of protons

= 35 - 17

= 18 neutrons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$atomic number

= no. of protons

= 17

$\green{\rule{5.2cm}{0.03cm}}$

$\large \boxed{3}$

Given :-

• L shell is completely filled.
• cation : (divalent)
• mass number , A = 24

To Find :-

$\bigstar$no. of electrons in cation

= 10 electrons

$\red{\rule{3.4cm}{0.03cm}}$

no. of electrons

= (electrons upto L-shell) + 2

= (2n$_{1} ^{2}$ + 2n) + 2

$\qquad$ [ here, n$_2$ is 2 for L-shell and n$_1$ 1 for K-shell ]

= 10 + 2

= 12 electrons

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$\bigstar$no. of protons

= no. of electrons

= 12 protons

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$\bigstar$no. of neutrons

= A - no. of protons

= 24 - 12

= 12 neutrons

$\green{\rule{5.2cm}{0.03cm}}$

$\large \boxed{4}$

Given :-

• L-shell is completely filled.
• anion : (monovalent)
• mass number , A = 19

To Find :-

$\bigstar$no. of electrons

= (electrons upto L-shell) - 1

= (2n$_{1} ^{2}$ + 2n) - 1

$\qquad$ [ here, n$_2$ is 2 for L-shell and n$_1$ 1 for K-shell ]

= 10 - 1

= 9 electrons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$no. of protons

= no. of electrons

= 9 protons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$no. of neutrons

= A - no. of protons

= 19 - 9

= 10 neutrons

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$atomic number

= no. of protons

= 9

$\red{\rule{3.4cm}{0.03cm}}$

$\bigstar$electronic configuration = ( 2 , 7 )

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