A copper wire of resistance 12 ohm. Its bent in the form of a circle. The effective resistance b/w two points a across a diameter is
a) 3 ohm
b) 6 ohm
c) 12 ohm
d) 24 ohm
For Solution see the pic:
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An element that would form anion of largest size to get a noble gas electronic configuration is :
This question tests your knowledge of cations and anions and their sizes in comparison to their parent atoms.
Remember the two points:
1) Size of cation is always smaller than the parent atom
2) Size of anion is always larger than the parent atom.
Now check among the above given options which one will form anion and which one will form cation to gain the noble gas electronic configuration.
a) Sodium: Its oxidation state is +1 and thus will form a cation.
b) phosphorous: Its oxidation states are +3 and +5 and thus will form cation.
c) chlorine: its oxidation state is -1 and thus will form anion.
d) sulphur: its oxidation state is -2 and will form anion.
For sodium and phosphorous the formation of anion isn't possible to get the noble gas electronic configuration so we are left with two options now.
Moving across the periods size of an atom decreases, so size of Chlorine atom is smaller than that of Sulfur. Moreover Chlorine gets one electron to complete the octet while sulphur gets two electrons. So more the electrons more the repulsion and thus greater the size..
Therefore, the answer to this question is sulphur.
Hope its clear now :) In case of any confusion fee free to ask...