Chemistry-
General
Easy

Question

Electron affinity depends on

  1. Atomic size  
  2. Nuclear charge   
  3. Atomic number   
  4. Atomic size and nuclear charge both  

The correct answer is: Atomic size and nuclear charge both

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The formation of the oxide ion O subscript left parenthesis g right parenthesis end subscript superscript 2 minus end superscript requires first an exothermic and then an endothermic step as shown below
O subscript left parenthesis g right parenthesis end subscript plus e to the power of minus end exponent equals O subscript left parenthesis g right parenthesis end subscript superscript minus end superscript capital delta H to the power of 0 end exponent equals negative 142   k J m o l to the power of negative 1 end exponent
O subscript left parenthesis g right parenthesis end subscript superscript minus end superscript plus e to the power of minus end exponent equals O subscript left parenthesis g right parenthesis end subscript superscript 2 minus end superscript capital delta H to the power of 0 end exponent equals 844   k J m o l to the power of negative 1 end exponent This is because

The formation of the oxide ion O subscript left parenthesis g right parenthesis end subscript superscript 2 minus end superscript requires first an exothermic and then an endothermic step as shown below
O subscript left parenthesis g right parenthesis end subscript plus e to the power of minus end exponent equals O subscript left parenthesis g right parenthesis end subscript superscript minus end superscript capital delta H to the power of 0 end exponent equals negative 142   k J m o l to the power of negative 1 end exponent
O subscript left parenthesis g right parenthesis end subscript superscript minus end superscript plus e to the power of minus end exponent equals O subscript left parenthesis g right parenthesis end subscript superscript 2 minus end superscript capital delta H to the power of 0 end exponent equals 844   k J m o l to the power of negative 1 end exponent This is because

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If the IP of N a is 5.48 eV, the ionisation potential of K will be

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