Position of element in periodic table is depend on the electronic configuration of element.
Element with 62 electrons has following electronic configuration:
<span>1s2 2s2 </span>2p6 <span>3s2 </span>3p6 4s2 3d10 4p6 <span>5s2 </span>4d10 5p6 4f6 <span>6s<span>2
</span></span>
From above electronic configuration, it can be seen that highest value of principal quantum number, where electron is present, is 6. Hence, element belongs to 6th period.
Further, last electron has entered f-orbital, hence it is a f-block element. Position of f-block element is the bottom of periodic table.
Further, there are 6 electrons in f-orbital. Hence, it is the 6th f-block element in 6th period of periodic table.
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Answer:
Kc = 168.0749
Explanation:
initial mol: 0.822 0 0
equil. mol: 2(0.822 - x) x x
∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )
⇒ 1.644 - 2x = 0.055 * 1.11
⇒ 1.644 = 2x + 0.06105
⇒ 2x = 1.583
⇒ x = 0.7915 mol equilibrium
⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq
⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²
⇒ Kc = ( 0.7130² ) / ( 0.055² )
⇒ Kc = 168.0749
