Question

An element ‘a' belongs to 3rd period and 17th group of the periodic table.

a.write atomic member and electronic configuration of ‘a'

b.state whether a is a metal or a non – metal.

c.write the nature of bond formed if a reacts with another element b of electronic configuration 2, 8, 1.

d.write the chemical formula of the compound formed.

Answer 1

a. atomic number is 17.
K L M
2 8 7
b. ' a ' is a non metal.
c. ionic bond .
element ' b ' is sodium with atomic number 11
d.NaCl

Answer 2

Answer: a) The atomic member of a is 35 and electronic configuration is $1s^22s^22p^63s^23p^63d^104s^24p^5$

b) Bromine is a non metal as it accepts electrons to acquire noble gas configuration. It is short of one electron from its nearest noble gas krypton which has 36 electrons.

c) The nature of the bond will be ionic.

$Br:1s^22s^22p^63s^23p^63d^104s^24p^5$

$Na :1s^22s^22p^63s^1$

Sodium $Na$ is a metal and lose its valence 1 electron to attain noble gas configuration to give $Na^+$. Bromine is a non metal and accepts 1 electron to attain noble gas configuration to give $Br^-$. Thus there is complete transfer of electrons from [ytex]Na[/tex] to $Br$ and thus form ionic bond.

d) The chemical formula of the compound is $NaBr$. To form a neutral compound one $Na^+$ combines with one $Br^-$ to give $NaBr$.