The question is in complete, the complete question is:
Which element has an outer electron configuration of s
2
p4
?
(A) Ca (B) Cr (C) Ge (D) Se
Answer:
(D) Se
Explanation:
The outer electronic configuration of an atom reflects the group to which it belongs in the periodic table. Each group has its unique outermost electron configuration and number of electrons found in the outermost shell of the atoms of elements in the group.
s
2
p4 is the unique outermost electron configuration of group 16 elements. Group 16 elements have six electrons in their outermost shell. If we consider the options, only Se is a member of group 16. Hence it is only Se that has the outermost electron configuration s
2
p4
.
Answer:
attracted towards each other
Explanation:
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Due to electronegativity difference between H and F atom.
<h3>
What is the reason for the difference ?</h3>
- The bond length of HF is 92 pm. It is 16% shorter than the sum of the covalent radii of H (37 pm) and F (72 pm).
- The shorter of the bond length is due to the difference in electronegativity between H and F atom. The electronegativity of F is 4.0, whereas the electronegativity of H is 2.1.
- Greater the difference of electronegativity between the bonded atoms, shorter will be the bond length.
- As becomes smaller i.e. as we proceed from HF to HI, this effect becomes lessens and bond length will be more predictable.
Thus from the above conclusion we can say the difference in electronegativity between H and F atom this is the reason for shorter bond length of HF.
Learn more about bond length here :
brainly.com/question/13683866
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<span>The
equilibrium constant is a value which represents the equilibrium of a reaction.
It is a reaction quotient when the reaction reached equilibrium. If Keq is
greater than 1, the mixture contains mostly the products. On the other hand, if
Keq is less than 1, the mixture contains the reactants. For this case, the equilibrium is towards the reactants or to the left.</span>