Consider the following electron configurations to answer the questions that follow:
(i) 1s2 2s2 2p6 3s1
(ii) 1s2 2s2 2p6 3s2
(iii) 1s2 2s2 2p6 3s2 3p1
(iv) 1s2 2s2 2p6 3s2 3p4
(v) 1s2 2s2 2p6 3s2 3p5
the electron configuration belonging to the atom with the highest second ionization energy is socratic
a) (i)
b) (ii)
c) (iii)
d) (iv)
e) (v)
the electron configuration that belongs to the atom with the lowest ionization energy is
a) (i)
b) (ii)
c) (iii)
d) (iv)
e) (v)

The correct options are: (A) (i) and (A) (i)

Explanation:

Ionization energy refers to the energy needed to completely remov valence electron from valence shell of the isolated gaseous atom.

First and the second ionization energy is the energy involved in the removal of first and the subsequent second electron.

The ionization energy is low when the removal of electron results in a stable electronic configuration (s²p⁶) and high if it disturbs the stable noble gas configuration (s²p⁶).

Among the given options, the electronic configuration: (i) 1s² 2s² 2p⁶ 3s¹, has the highest second ionization energy and the lowest ionization energy.

This is because the removal of first electron from the electronic configuration (1s² 2s² 2p⁶ 3s¹) is easy, because it results in the formation of stable noble gas configuration (1s² 2s² 2p⁶). Therefore, the first ionization is the lowest.

Further, the removal of second electron is very difficult, because it disturbs the stable electronic configuration (1s² 2s² 2p⁶) and results in the formation of unstable electron configuration (1s² 2s² 2p⁵). Therefore, the second ionization is highest.

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol