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SSSSS [86.1K]
2 years ago
12

Consider the following electron configurations to answer the questions that follow: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (ii

i) 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 ________.
Chemistry
1 answer:
Ahat [919]2 years ago
6 0

Option (i) would have the highest 2nd Ionization Energy.

Option (i) is Sodium.

Can be Written as 2, 8 , 1

For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.

Now After Removing that Electron...

Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.

Neon has a very stable Octet.

It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.

So

Option (i) [Sodium] has the highest 2nd Ionization Energy

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Answer:

4.993 ×10⁻¹¹ J

Explanation:

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The <em>mass defect</em> is the difference between the mass of a nucleus and the sum of the masses of its nucleons.

<em>Calculate the mass defect </em>

  16 p = 16 × 1.007 28 u =  16.116 48  u

  16 n = 16 × 1.008 67 u =  16.138 72  u

Total mass of nucleons = 32.255 20 u

             - Mass of S-32 = <u>31.972 070 u </u>

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Convert the <em>unified atomic mass units to kilograms</em>.

Mass defect

= \text{0.283 13 u} \times \frac{ \text{1.66 06} \times 10^{-27} \text{ kg}}{\text{1 u}}

= 4.700 \times 10^{-28} \text{ kg}

Use Einstein’s equation to <em>convert the mass defect into energy</em>

E = mc^{2}

E = 4.7000 \times 10^{-28} \text{ kg} \times (2.998 \times 10^{8} \text{ m}\cdot\text{s}^{-1})^{2} = 4.224 \times 10^{-11} \text{ J}

4 0
3 years ago
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acid

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<span>B</span>

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