Answer:
Sulfur: -1
Carbon: 0
Nitrogen: 0
Explanation:
The thiocyanate ion SCN- can have only two resonance structures, which are:
S - C ≡ N <--------> S = C = N
In the first structure, we have one single bond and one triple bond, in this case, the negative charge is located in the sulfur. This is because Sulfur have 6 electrons and those electrons are present in the atom, (see picture below), and counting the electron that is sharing with the Carbon, the total electrons that sulfur has is 7 (It has one more than usual). Carbon and nitrogen are already stable with 0 of formal charge, because carbon can only have 4 electrons which 1 is sharing with sulfur and the other 3 with the nitrogen, and nitrogen have 5 electrons, three sharing with carbon and the other two kept it for itself.
In the second structure, the negative charge of the sulfur is transfered to the nitrogen, meaning that it has 6 electrons the nitrogen (formal charge -1) and carbon and sulfur with 4 and 6 electrons respectively.
Between these two structures, the most stable is the first one basically because Sulfur is a better nucleophile than the Nitrogen, and can form stronger hydrogen bond in acid, giving more stable structure.
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).
2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.
3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.
4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
I would say b and d but I wouldn’t really know :/
The molecular formula of the given compound is CH₅N.
<h3>What is the ideal gas equation?</h3>
The ideal gas law has defined the product of the volume and pressure as equal to the product of the gas constant (R) and absolute temperature of the gas.
The mathematical equation of the ideal gas law is as follows:
PV = nRT
The temperature of the compound, T = 22° C = 295 K
The pressure of the compound, P = 99.5 KPa = 0.982 atm
The volume of the compound, V = 125 ml = 0.125 L
0.125 × 0.982 = (0.157 g/M) × 0.082 ×295
M = 31 g/mol
A number of moles of C = 0.157 ×0.37/12 = 0.00484
A number of moles of N = 0.157 ×0.45/14 = 0.00504
A number of moles of H = 0.157 ×0.16/1 = 0.0251
The simplest ratio of C: N: H is 1 : 1 : 5
The empirical formula of the given compound is CH₅N.
The molecular formula of the compound is CH₅N as the molecular mass is equal to 31 g/mol.
Learn more about ideal gas equation, here:
brainly.com/question/3637553
#SPJ1
First identify which is being oxidized and reduced. In this case, the Mg is being oxidized and the Hg is being reduced.
Mg --> Mg+2
<span>Hg+2 --> Hg+1
</span>
Then you have to balance each half reaction first with electrons before adding them together in one equation
⇒
and
⇒
and then combine them together to form
⇒
It isn't necessary to keep the electrons but its essential to know how many there are in order to know how many are in the equation in order to calculate the reaction energy. Note: A<span>dd H+ and H2O to balance the H's and O's in acidic solution if needed.</span>
