As we know that wavelength and frequency is inversely proportional to each other. Greater the wavelength smaller the frequency and vice versa.
Solution:
The relation between wavelength and frequency is as follow,
υ = c / λ
where
υ = frequency = ?
c = velocity of light = 3.0 × 10⁸ ms⁻¹
λ = wavenumber = 542 nm = 542 × 10⁻⁹ m
Putting the given values,
υ = 3.0 × 10⁸ ms⁻¹ / 542 × 10⁻⁹ m
Result:
υ = 5.53 × 10¹⁴ s⁻¹
Answer:
Have 2 filled orbitals and 3 partially filled orbitals.
Explanation:
Hello there!
In this case, according to the given information of the electron configuration for that particle; it is possible for us to infer it has 5 valence electrons, as the electrons on its outermost shell (2). Moreover, we undertand this particle needs three bonds, does not have neither the electron configuration of a noble gas which ends by p⁶ nor that of an alkali earth metal as it ends by s².
Therefore, we infer the correct answer is Have 2 filled orbitals and 3 partially filled orbitals because according to the Hund's rule, the s orbital is fulfilled and the p orbital has 1 electron orbital fulfilled and two partially filled orbitals.
Regards!
The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au
