It would have a zero charge because it is a neutral atom. The number of electrons which is a negative charge is equal to the number of protons which is positive, they will cancel each other out hence meaning it will become neutral
11 with pemdas, you have to multiply all by 3
Yes they are stable because they follow octet rule but am not sure if they are exist separetly
I don't see the options for an answer, so here is a list of all of the transition metals lol
- <em>Scandium</em>
- <em>Titanium</em>
- <em>Vanadium</em>
- <em>Chromium</em>
- <em>Manganese</em>
- <em>Iron</em>
- <em>Cobalt</em>
- <em>Nickel</em>
- <em>Copper</em>
- <em>Zinc</em>
- <em>Yttrium</em>
- <em>Zirconium</em>
- <em>Niobium</em>
- <em>Molybdenum</em>
- <em>Technetium</em>
- <em>Ruthenium</em>
- <em>Rhodium</em>
- <em>Palladium</em>
- <em>Silver</em>
- <em>Cadmium</em>
- <em>Lanthanum</em>
- <em>Hafnium</em>
- <em>Tantalum</em>
- <em>Tungsten</em>
- <em>Rhenium</em>
- <em>Osmium</em>
- <em>Iridium</em>
- <em>Platinum</em>
- <em>Gold</em>
- <em>Mercury</em>
- <em>Actinium</em>
- <em>Rutherfordium</em>
- <em>Dubnium</em>
- <em>Seaborgium</em>
- <em>Bohrium</em>
- <em>Hassium</em>
- <em>Meitnerium</em>
- <em>Darmstadtium</em>
- <em>Roentgenium</em>
- <em>Copernicium p</em>
Aromatic compound has continuous cyclic structure with( 4n+2)π electrons (Huckels rule), where n = 0,1,2…
Here number of pi electron are 6, where 4 from two double bond and 2 from nitrogen non-bonding electrons, hence it has total 6 pi electrons, therefore
6= ( 4n+2)π
4 = 4n
n =1
Hence it is an aromatic compound
